Chapter 14 from Textbook Rs Aggarwal 2021 2022 for Class 10 MATHS

Question 1:

Answer:

Let $A B$ be the tower standing vertically on the ground and O be the position of the observer.
We now have:
$O A = 20 m,$ ∠ $O A B = 90^{o}$ and ∠ $A O B$ $= 60^{o}$
Let:
$A B = h$ m

Now, in the right âˆ† $O A B$ , we have:
$\frac{A B}{O A} = \tan 60^{o}$ = $\sqrt{3}$

⇒ $\frac{h}{20} = \sqrt{3}$

⇒ $h = 20 \sqrt{3}$ = $(20 \times 1.732) = 34.64$

Hence, the height of the pole is 34.64 m.

Question 2:

Let $A B$ be the tower standing vertically on the ground and O be the position of the observer.
We now have:
$O A = 20 m,$ ∠ $O A B = 90^{o}$ and ∠ $A O B$ $= 60^{o}$
Let:
$A B = h$ m

Now, in the right âˆ† $O A B$ , we have:
$\frac{A B}{O A} = \tan 60^{o}$ = $\sqrt{3}$

⇒ $\frac{h}{20} = \sqrt{3}$

⇒ $h = 20 \sqrt{3}$ = $(20 \times 1.732) = 34.64$

Hence, the height of the pole is 34.64 m.

Answer:

Let $O X$ be the horizontal ground and $A$ be the position of the kite.
Also, let O be the position of the observer and $O A$ be the thread.
Now, draw $A B$ ⊥ $O X$ .
We have:
∠ $B O A =$ $60^{o}$ , $O A = 75$ m and ∠ $O B A = 90^{o}$
Height of the kite from the ground = $A B$ = 75 m
Length of the string, $O A = x$ m

In the right âˆ† $O B A$ , we have:
$\frac{A B}{O A} = \sin 60^{o} = \frac{\sqrt{3}}{2}$
⇒ $\frac{75}{x} = \frac{\sqrt{3}}{2}$
⇒ $x = \frac{75 \times 2}{\sqrt{3}} = \frac{150}{1.732} = 86.6$ m
Hence, the length of the string is $86.6$ m.

Question 3:

Let $O X$ be the horizontal ground and $A$ be the position of the kite.
Also, let O be the position of the observer and $O A$ be the thread.
Now, draw $A B$ ⊥ $O X$ .
We have:
∠ $B O A =$ $60^{o}$ , $O A = 75$ m and ∠ $O B A = 90^{o}$
Height of the kite from the ground = $A B$ = 75 m
Length of the string, $O A = x$ m

In the right âˆ† $O B A$ , we have:
$\frac{A B}{O A} = \sin 60^{o} = \frac{\sqrt{3}}{2}$
⇒ $\frac{75}{x} = \frac{\sqrt{3}}{2}$
⇒ $x = \frac{75 \times 2}{\sqrt{3}} = \frac{150}{1.732} = 86.6$ m
Hence, the length of the string is $86.6$ m.

Answer:

Let CE and AD be the heights of the observer and the chimney, respectively.

We have,
$BD = CE = 1.5 m, BC = DE = 30 m and ∠ ACB = 60 ° In ∆ ABC, \tan 60 ° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{AD - BD}{30} \Rightarrow AD - 1.5 = 30 \sqrt{3} \Rightarrow AD = 30 \sqrt{3} + 1.5 \Rightarrow AD = 30 \times 1.732 + 1.5 \Rightarrow AD = 51.96 + 1.5 \Rightarrow AD = 53.46 m$

So, the height of the chimney is 53.46 m (approx.).

Question 4:

Let CE and AD be the heights of the observer and the chimney, respectively.

We have,
$BD = CE = 1.5 m, BC = DE = 30 m and ∠ ACB = 60 ° In ∆ ABC, \tan 60 ° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{AD - BD}{30} \Rightarrow AD - 1.5 = 30 \sqrt{3} \Rightarrow AD = 30 \sqrt{3} + 1.5 \Rightarrow AD = 30 \times 1.732 + 1.5 \Rightarrow AD = 51.96 + 1.5 \Rightarrow AD = 53.46 m$

So, the height of the chimney is 53.46 m (approx.).

Answer:

Let the height of the tower be AB.

We have,
$AC = 5 m, AD = 20 m Let the angle of elevation of the top of the tower (i . e . ∠ ACB) from point C be θ . Then, the angle of elevation of the top of the tower (i . e . ∠ ADB) from point D = (90 ° - θ) Now, in ∆ ABC, \tan θ = \frac{AB}{AC} \Rightarrow \tan θ = \frac{AB}{5} . . . . . (i) Also, in ∆ ABD, \cot (90 ° - θ) = \frac{AD}{AB} \Rightarrow \tan θ = \frac{20}{AB} . . . . . (ii) From (i) and (ii), we get \frac{AB}{5} = \frac{20}{AB} \Rightarrow {AB}^{2} = 100 \Rightarrow AB = \sqrt{100} ∴ AB = 10 m$

So, the height of the tower is 10 m.

Question 5:

Let the height of the tower be AB.

We have,
$AC = 5 m, AD = 20 m Let the angle of elevation of the top of the tower (i . e . ∠ ACB) from point C be θ . Then, the angle of elevation of the top of the tower (i . e . ∠ ADB) from point D = (90 ° - θ) Now, in ∆ ABC, \tan θ = \frac{AB}{AC} \Rightarrow \tan θ = \frac{AB}{5} . . . . . (i) Also, in ∆ ABD, \cot (90 ° - θ) = \frac{AD}{AB} \Rightarrow \tan θ = \frac{20}{AB} . . . . . (ii) From (i) and (ii), we get \frac{AB}{5} = \frac{20}{AB} \Rightarrow {AB}^{2} = 100 \Rightarrow AB = \sqrt{100} ∴ AB = 10 m$

So, the height of the tower is 10 m.

Answer:

Let BC and CD be the heights of the tower and the flagstaff, respectively.

We have,
$AB = 120 m, ∠ BAC = 45 °, ∠ BAD = 60 ° Let CD = x In ∆ ABC, \tan 45 ° = \frac{BC}{AB} \Rightarrow 1 = \frac{BC}{120} \Rightarrow BC = 120 m Now, in ∆ ABD, \tan 60 ° = \frac{BD}{AB} \Rightarrow \sqrt{3} = \frac{BC + CD}{120} \Rightarrow BC + CD = 120 \sqrt{3} \Rightarrow 120 + x = 120 \sqrt{3} \Rightarrow x = 120 \sqrt{3} - 120 \Rightarrow x = 120 (\sqrt{3} - 1) \Rightarrow x = 120 (1.732 - 1) \Rightarrow x = 120 (0.732) \Rightarrow x = 87.84 \approx 87.8 m$

So, the height of the flagstaff is 87.8 m.

Question 6:

Let BC and CD be the heights of the tower and the flagstaff, respectively.

We have,
$AB = 120 m, ∠ BAC = 45 °, ∠ BAD = 60 ° Let CD = x In ∆ ABC, \tan 45 ° = \frac{BC}{AB} \Rightarrow 1 = \frac{BC}{120} \Rightarrow BC = 120 m Now, in ∆ ABD, \tan 60 ° = \frac{BD}{AB} \Rightarrow \sqrt{3} = \frac{BC + CD}{120} \Rightarrow BC + CD = 120 \sqrt{3} \Rightarrow 120 + x = 120 \sqrt{3} \Rightarrow x = 120 \sqrt{3} - 120 \Rightarrow x = 120 (\sqrt{3} - 1) \Rightarrow x = 120 (1.732 - 1) \Rightarrow x = 120 (0.732) \Rightarrow x = 87.84 \approx 87.8 m$

So, the height of the flagstaff is 87.8 m.

Answer:

Let BC be the tower and CD be the water tank.

$We have, AB = 40 m, ∠ BAC = 30 ° and ∠ BAD = 45 ° In ∆ ABD, \tan 45 ° = \frac{BD}{AB} \Rightarrow 1 = \frac{BD}{40} \Rightarrow BD = 40 m Now, in ∆ ABC, \tan 30 ° = \frac{BC}{AB} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{40} \Rightarrow BC = \frac{40}{\sqrt{3}} \Rightarrow BC = \frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow BC = \frac{40 \sqrt{3}}{3} m (i) The height of the tower, BC = \frac{40 \sqrt{3}}{3} = \frac{40 \times 1.73}{3} = 23.067 \approx 23.1 m (ii) The depth of the tank, CD = (BD - BC) = (40 - 23.1) = 16.9 m$

Question 7:

Let BC be the tower and CD be the water tank.

$We have, AB = 40 m, ∠ BAC = 30 ° and ∠ BAD = 45 ° In ∆ ABD, \tan 45 ° = \frac{BD}{AB} \Rightarrow 1 = \frac{BD}{40} \Rightarrow BD = 40 m Now, in ∆ ABC, \tan 30 ° = \frac{BC}{AB} \Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{40} \Rightarrow BC = \frac{40}{\sqrt{3}} \Rightarrow BC = \frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow BC = \frac{40 \sqrt{3}}{3} m (i) The height of the tower, BC = \frac{40 \sqrt{3}}{3} = \frac{40 \times 1.73}{3} = 23.067 \approx 23.1 m (ii) The depth of the tank, CD = (BD - BC) = (40 - 23.1) = 16.9 m$

Answer:

Let AB be the tower and BC be the flagstaff.

We have,
$BC = 6 m, ∠ AOB = 30 ° and ∠ AOC = 60 ° Let AB = h In ∆ AOB, \tan 30 ° = \frac{AB}{OA} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{OA} \Rightarrow OA = h \sqrt{3} . . . . . (i) Now, in ∆ AOC, \tan 60 ° = \frac{AC}{OA} \Rightarrow \sqrt{3} = \frac{AB + BC}{h \sqrt{3}} [Using (i)] \Rightarrow 3 h = h + 6 \Rightarrow 3 h - h = 6 \Rightarrow 2 h = 6 \Rightarrow h = \frac{6}{2} \Rightarrow h = 3 m$

So, the height of the tower is 3 m.

Question 8:

Let AB be the tower and BC be the flagstaff.

We have,
$BC = 6 m, ∠ AOB = 30 ° and ∠ AOC = 60 ° Let AB = h In ∆ AOB, \tan 30 ° = \frac{AB}{OA} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{OA} \Rightarrow OA = h \sqrt{3} . . . . . (i) Now, in ∆ AOC, \tan 60 ° = \frac{AC}{OA} \Rightarrow \sqrt{3} = \frac{AB + BC}{h \sqrt{3}} [Using (i)] \Rightarrow 3 h = h + 6 \Rightarrow 3 h - h = 6 \Rightarrow 2 h = 6 \Rightarrow h = \frac{6}{2} \Rightarrow h = 3 m$

So, the height of the tower is 3 m.

Answer:

Let $A C$ be the pedestal and $B C$ be the statue such that $B C = 1.46$ m.
We have:
∠ $A D C = 45^{o}$ and ∠ $A D B = 60^{o}$
Let:
$A C = h$ m and $A D = x$ m

In the right âˆ† $A D C$ , we have:
$\frac{A C}{A D} = \tan 45^{o} = 1$

⇒ $\frac{h}{x} = 1$
⇒ $h = x$
Or,
$x = h$

Now, in the right âˆ† $A D B$ , we have:
$\frac{A B}{A D} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{h + 1.46}{x} = \sqrt{3}$

On putting $x = h$ in the above equation, we get:
$\frac{h + 1.46}{h} = \sqrt{3}$

⇒ $h + 1.46 = \sqrt{3} h$
⇒ $h (\sqrt{3} - 1) = 1.46$
⇒ $h = \frac{1.46}{(\sqrt{3} - 1)} = \frac{1.46}{0.73} = 2$ m

Hence, the height of the pedestal is 2 m.

Question 9:

Let $A C$ be the pedestal and $B C$ be the statue such that $B C = 1.46$ m.
We have:
∠ $A D C = 45^{o}$ and ∠ $A D B = 60^{o}$
Let:
$A C = h$ m and $A D = x$ m

In the right âˆ† $A D C$ , we have:
$\frac{A C}{A D} = \tan 45^{o} = 1$

⇒ $\frac{h}{x} = 1$
⇒ $h = x$
Or,
$x = h$

Now, in the right âˆ† $A D B$ , we have:
$\frac{A B}{A D} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{h + 1.46}{x} = \sqrt{3}$

On putting $x = h$ in the above equation, we get:
$\frac{h + 1.46}{h} = \sqrt{3}$

⇒ $h + 1.46 = \sqrt{3} h$
⇒ $h (\sqrt{3} - 1) = 1.46$
⇒ $h = \frac{1.46}{(\sqrt{3} - 1)} = \frac{1.46}{0.73} = 2$ m

Hence, the height of the pedestal is 2 m.

Answer:

Let $A B$ be the unfinished tower, $A C$ be the raised tower and O be the point of observation.
We have:
$O A = 75$ m, ∠ $A O B = 30^{o}$ and ∠ $A O C = 60^{o}$
Let $A C = H$ m such that $B C = (H - h)$ m.

In âˆ†AOB, we have:
$\frac{A B}{O A} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{75} = \frac{1}{\sqrt{3}}$

⇒ $h = \frac{75}{\sqrt{3}}$ m = $\frac{75 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = 25 \sqrt{3}$ m
In âˆ† $A O C$ , we have:
$\frac{A C}{O A} = \tan 60^{o}$ $= \sqrt{3}$

⇒ $\frac{H}{75} = \sqrt{3}$
⇒ $H = 75 \sqrt{3}$ m

∴ Required height = $(H - h) = (75 \sqrt{3} - 25 \sqrt{3}) = 50 \sqrt{3}$ m = 86.6 m

Question 10:

Let $A B$ be the unfinished tower, $A C$ be the raised tower and O be the point of observation.
We have:
$O A = 75$ m, ∠ $A O B = 30^{o}$ and ∠ $A O C = 60^{o}$
Let $A C = H$ m such that $B C = (H - h)$ m.

In âˆ†AOB, we have:
$\frac{A B}{O A} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{75} = \frac{1}{\sqrt{3}}$

⇒ $h = \frac{75}{\sqrt{3}}$ m = $\frac{75 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = 25 \sqrt{3}$ m
In âˆ† $A O C$ , we have:
$\frac{A C}{O A} = \tan 60^{o}$ $= \sqrt{3}$

⇒ $\frac{H}{75} = \sqrt{3}$
⇒ $H = 75 \sqrt{3}$ m

∴ Required height = $(H - h) = (75 \sqrt{3} - 25 \sqrt{3}) = 50 \sqrt{3}$ m = 86.6 m

Answer:

Let $O X$ be the horizontal plane, $A D$ be the tower and $C D$ be the vertical flagpole.
We have:
$A B = 9$ m, ∠ $D B A = 30^{o}$ and ∠ $C B A = 60^{o}$
Let:
$A D = h$ m and $C D = x$ m

In the right âˆ† $A B D$ , we have:
$\frac{A D}{A B} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{h}{9} = \frac{1}{\sqrt{3}}$
⇒ $h = \frac{9}{\sqrt{3}} = 5.19$ m
Now, in the right âˆ† $A B C$ , we have:
$\frac{A C}{B A} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{h + x}{9} = \sqrt{3}$
⇒ $h + x = 9 \sqrt{3}$

By putting $h = \frac{9}{\sqrt{3}}$ in the above equation, we get:
$\frac{9}{\sqrt{3}} + x = 9 \sqrt{3}$
⇒ $x = 9 \sqrt{3} - \frac{9}{\sqrt{3}}$
⇒ $x = \frac{27 - 9}{\sqrt{3}} = \frac{18}{\sqrt{3}} = \frac{18}{1.73} = 10.4$

Thus, we have:
Height of the flagpole = 10.4 m
Height of the tower = 5.19 m

Question 11:

Let $O X$ be the horizontal plane, $A D$ be the tower and $C D$ be the vertical flagpole.
We have:
$A B = 9$ m, ∠ $D B A = 30^{o}$ and ∠ $C B A = 60^{o}$
Let:
$A D = h$ m and $C D = x$ m

In the right âˆ† $A B D$ , we have:
$\frac{A D}{A B} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{h}{9} = \frac{1}{\sqrt{3}}$
⇒ $h = \frac{9}{\sqrt{3}} = 5.19$ m
Now, in the right âˆ† $A B C$ , we have:
$\frac{A C}{B A} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{h + x}{9} = \sqrt{3}$
⇒ $h + x = 9 \sqrt{3}$

By putting $h = \frac{9}{\sqrt{3}}$ in the above equation, we get:
$\frac{9}{\sqrt{3}} + x = 9 \sqrt{3}$
⇒ $x = 9 \sqrt{3} - \frac{9}{\sqrt{3}}$
⇒ $x = \frac{27 - 9}{\sqrt{3}} = \frac{18}{\sqrt{3}} = \frac{18}{1.73} = 10.4$

Thus, we have:
Height of the flagpole = 10.4 m
Height of the tower = 5.19 m

Answer:

Let AB and CD be the equal poles; and BD be the width of the road.

We have,
$∠ AOB = 60 ° and ∠ COD = 60 ° In ∆ AOB, \tan 60 ° = \frac{AB}{BO} \Rightarrow \sqrt{3} = \frac{AB}{BO} \Rightarrow BO = \frac{AB}{\sqrt{3}} Also, in ∆ COD, \tan 30 ° = \frac{CD}{DO} \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{DO} \Rightarrow DO = \sqrt{3} CD As, BD = 80 \Rightarrow BO + DO = 80 \Rightarrow \frac{AB}{\sqrt{3}} + \sqrt{3} CD = 80 \Rightarrow \frac{AB}{\sqrt{3}} + \sqrt{3} AB = 80 (Given : AB = CD) \Rightarrow AB (\frac{1}{\sqrt{3}} + \sqrt{3}) = 80 \Rightarrow AB (\frac{1 + 3}{\sqrt{3}}) = 80 \Rightarrow AB (\frac{4}{\sqrt{3}}) = 80 \Rightarrow AB = \frac{80 \sqrt{3}}{4} \Rightarrow AB = 20 \sqrt{3} m Also, BO = \frac{AB}{\sqrt{3}} = \frac{20 \sqrt{3}}{\sqrt{3}} = 20 m So, DO = 80 - 20 = 60 m$

Hence, the height of each pole is 20 $\sqrt{3}$ m and point P is at a distance of 20 m from left pole and 60 m from right pole.

Question 12:

Let AB and CD be the equal poles; and BD be the width of the road.

We have,
$∠ AOB = 60 ° and ∠ COD = 60 ° In ∆ AOB, \tan 60 ° = \frac{AB}{BO} \Rightarrow \sqrt{3} = \frac{AB}{BO} \Rightarrow BO = \frac{AB}{\sqrt{3}} Also, in ∆ COD, \tan 30 ° = \frac{CD}{DO} \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{DO} \Rightarrow DO = \sqrt{3} CD As, BD = 80 \Rightarrow BO + DO = 80 \Rightarrow \frac{AB}{\sqrt{3}} + \sqrt{3} CD = 80 \Rightarrow \frac{AB}{\sqrt{3}} + \sqrt{3} AB = 80 (Given : AB = CD) \Rightarrow AB (\frac{1}{\sqrt{3}} + \sqrt{3}) = 80 \Rightarrow AB (\frac{1 + 3}{\sqrt{3}}) = 80 \Rightarrow AB (\frac{4}{\sqrt{3}}) = 80 \Rightarrow AB = \frac{80 \sqrt{3}}{4} \Rightarrow AB = 20 \sqrt{3} m Also, BO = \frac{AB}{\sqrt{3}} = \frac{20 \sqrt{3}}{\sqrt{3}} = 20 m So, DO = 80 - 20 = 60 m$

Hence, the height of each pole is 20 $\sqrt{3}$ m and point P is at a distance of 20 m from left pole and 60 m from right pole.

Answer:

Let $C D$ be the tower and $A and B$ be the positions of the two men standing on the opposite sides. Thus, we have:
∠ $D A C = 30^{o}$ , ∠ $D B C = 45^{o}$ and $C D = 50$ m
Let $A B = x$ m and $B C = y$ m such that $A C = (x - y)$ m.

In the right âˆ† $D B C$ , we have:
$\frac{C D}{B C} = \tan 45^{o} = 1$

⇒ $\frac{50}{y} = 1$
⇒ $y = 50$ m

In the right âˆ† $A C D$ , we have:
$\frac{C D}{A C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{50}{(x - y)} = \frac{1}{\sqrt{3}}$
⇒ $x - y = 50 \sqrt{3}$
On putting $y = 50$ in the above equation, we get:
$x - 50 = 50 \sqrt{3}$
⇒ $x = 50 + 50 \sqrt{3} = 50 (\sqrt{3} + 1) = 136.6$ m

∴ Distance between the two men = $A B = x = 136.6$ m

Question 13:

Let $C D$ be the tower and $A and B$ be the positions of the two men standing on the opposite sides. Thus, we have:
∠ $D A C = 30^{o}$ , ∠ $D B C = 45^{o}$ and $C D = 50$ m
Let $A B = x$ m and $B C = y$ m such that $A C = (x - y)$ m.

In the right âˆ† $D B C$ , we have:
$\frac{C D}{B C} = \tan 45^{o} = 1$

⇒ $\frac{50}{y} = 1$
⇒ $y = 50$ m

In the right âˆ† $A C D$ , we have:
$\frac{C D}{A C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{50}{(x - y)} = \frac{1}{\sqrt{3}}$
⇒ $x - y = 50 \sqrt{3}$
On putting $y = 50$ in the above equation, we get:
$x - 50 = 50 \sqrt{3}$
⇒ $x = 50 + 50 \sqrt{3} = 50 (\sqrt{3} + 1) = 136.6$ m

∴ Distance between the two men = $A B = x = 136.6$ m

Answer:

Let PQ be the tower.

We have,

$PQ = 100 m, ∠ PAQ = 30 ° and ∠ PBQ = 45 ° In ∆ APQ, \tan 30 ° = \frac{PQ}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{AP} \Rightarrow AP = 100 \sqrt{3} m Als, in ∆ BPQ, \tan 45 ° = \frac{PQ}{BP} \Rightarrow 1 = \frac{100}{BP} \Rightarrow BP = 100 m Now, AB = AP + BP = 100 \sqrt{3} + 100 = 100 (\sqrt{3} + 1) = 100 \times (1.73 + 1) = 100 \times 2.73 = 273 m$

So, the distance between the cars is 273 m.

Question 14:

Let PQ be the tower.

We have,

$PQ = 100 m, ∠ PAQ = 30 ° and ∠ PBQ = 45 ° In ∆ APQ, \tan 30 ° = \frac{PQ}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{AP} \Rightarrow AP = 100 \sqrt{3} m Als, in ∆ BPQ, \tan 45 ° = \frac{PQ}{BP} \Rightarrow 1 = \frac{100}{BP} \Rightarrow BP = 100 m Now, AB = AP + BP = 100 \sqrt{3} + 100 = 100 (\sqrt{3} + 1) = 100 \times (1.73 + 1) = 100 \times 2.73 = 273 m$

So, the distance between the cars is 273 m.

Answer:

Let PQ be the tower.

$We have, ∠ PBQ = 60 ° and ∠ PAQ = 30 ° Let PQ = h, AB = x and BQ = y In ∆ APQ, \tan 30 ° = \frac{PQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + y} \Rightarrow x + y = h \sqrt{3} . . . . . (i) Also, in ∆ BPQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{y} \Rightarrow h = y \sqrt{3} . . . . . (ii) Substituting h = y \sqrt{3} in (i), we get x + y = \sqrt{3} (y \sqrt{3}) \Rightarrow x + y = 3 y \Rightarrow 3 y - y = x \Rightarrow 2 y = x \Rightarrow y = \frac{x}{2} As, speed of the car from A to B = \frac{AB}{6} = \frac{x}{6} units / \sec So, the time taken to reach the foot of the tower i . e . Q from B = \frac{BQ}{speed} = \frac{y}{(\frac{x}{6})} = \frac{(\frac{x}{2})}{(\frac{x}{6})} = \frac{6}{2} = 3 \sec$

So, the time taken to reach the foot of the tower from the given point is 3 seconds.

Question 15:

Let PQ be the tower.

$We have, ∠ PBQ = 60 ° and ∠ PAQ = 30 ° Let PQ = h, AB = x and BQ = y In ∆ APQ, \tan 30 ° = \frac{PQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + y} \Rightarrow x + y = h \sqrt{3} . . . . . (i) Also, in ∆ BPQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{y} \Rightarrow h = y \sqrt{3} . . . . . (ii) Substituting h = y \sqrt{3} in (i), we get x + y = \sqrt{3} (y \sqrt{3}) \Rightarrow x + y = 3 y \Rightarrow 3 y - y = x \Rightarrow 2 y = x \Rightarrow y = \frac{x}{2} As, speed of the car from A to B = \frac{AB}{6} = \frac{x}{6} units / \sec So, the time taken to reach the foot of the tower i . e . Q from B = \frac{BQ}{speed} = \frac{y}{(\frac{x}{6})} = \frac{(\frac{x}{2})}{(\frac{x}{6})} = \frac{6}{2} = 3 \sec$

So, the time taken to reach the foot of the tower from the given point is 3 seconds.

Answer:

Let PQ=h m be the height of the TV tower and BQ=x m be the width of the canal.

We have,

$AB = 20 m, ∠ PAQ = 30 °, ∠ PBQ = 60 °, BQ = x and PQ = h In ∆ PBQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x \sqrt{3} . . . . . (i) Again, in ∆ APQ, \tan 30 ° = \frac{PQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x \sqrt{3}}{20 + x} [Using (i)] \Rightarrow 3 x = 20 + x \Rightarrow 3 x - x = 20 \Rightarrow 2 x = 20 \Rightarrow x = \frac{20}{2} \Rightarrow x = 10 m Substituting x = 10 in (i), we get h = 10 \sqrt{3} m$

So, the height of the TV tower is $10 \sqrt{3} m$ and the width of the canal is 10 m.

Question 16:

Let PQ=h m be the height of the TV tower and BQ=x m be the width of the canal.

We have,

$AB = 20 m, ∠ PAQ = 30 °, ∠ PBQ = 60 °, BQ = x and PQ = h In ∆ PBQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x \sqrt{3} . . . . . (i) Again, in ∆ APQ, \tan 30 ° = \frac{PQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x \sqrt{3}}{20 + x} [Using (i)] \Rightarrow 3 x = 20 + x \Rightarrow 3 x - x = 20 \Rightarrow 2 x = 20 \Rightarrow x = \frac{20}{2} \Rightarrow x = 10 m Substituting x = 10 in (i), we get h = 10 \sqrt{3} m$

So, the height of the TV tower is $10 \sqrt{3} m$ and the width of the canal is 10 m.

Answer:

Let AB be the building and PQ be the tower.

We have,

$PQ = 60 m, ∠ APB = 30 °, ∠ PAQ = 60 ° In ∆ APQ, \tan 60 ° = \frac{PQ}{AP} \Rightarrow \sqrt{3} = \frac{60}{AP} \Rightarrow AP = \frac{60}{\sqrt{3}} \Rightarrow AP = \frac{60 \sqrt{3}}{3} \Rightarrow AP = 20 \sqrt{3} m Now, in ∆ ABP, \tan 30 ° = \frac{AB}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{20 \sqrt{3}} \Rightarrow AB = \frac{20 \sqrt{3}}{\sqrt{3}} ∴ AB = 20 m$

So, the height of the building is 20 m.

Question 17:

Let AB be the building and PQ be the tower.

We have,

$PQ = 60 m, ∠ APB = 30 °, ∠ PAQ = 60 ° In ∆ APQ, \tan 60 ° = \frac{PQ}{AP} \Rightarrow \sqrt{3} = \frac{60}{AP} \Rightarrow AP = \frac{60}{\sqrt{3}} \Rightarrow AP = \frac{60 \sqrt{3}}{3} \Rightarrow AP = 20 \sqrt{3} m Now, in ∆ ABP, \tan 30 ° = \frac{AB}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{20 \sqrt{3}} \Rightarrow AB = \frac{20 \sqrt{3}}{\sqrt{3}} ∴ AB = 20 m$

So, the height of the building is 20 m.

Answer:

Let $D E$ be the first tower and $A B$ be the second tower.
Now, $A B = 90$ m and $A D = 60$ m such that $C E = 60$ m and ∠ $B E C = 30^{o}$ .
Let $D E = h$ m such that $A C = h$ m and $B C = (90 - h)$ m.

In the right âˆ† $B C E$ , we have:
$\frac{B C}{C E} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{(90 - h)}{60} = \frac{1}{\sqrt{3}}$
⇒ $(90 - h) \sqrt{3} = 60$
⇒ $h \sqrt{3} = 90 \sqrt{3} - 60$
⇒ $h = 90 - \frac{60}{\sqrt{3}}$ = $90 - 34.64 = 55.36$ m
∴ Height of the first tower = $D E = h = 55.36$ m

Question 18:

Let $D E$ be the first tower and $A B$ be the second tower.
Now, $A B = 90$ m and $A D = 60$ m such that $C E = 60$ m and ∠ $B E C = 30^{o}$ .
Let $D E = h$ m such that $A C = h$ m and $B C = (90 - h)$ m.

In the right âˆ† $B C E$ , we have:
$\frac{B C}{C E} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{(90 - h)}{60} = \frac{1}{\sqrt{3}}$
⇒ $(90 - h) \sqrt{3} = 60$
⇒ $h \sqrt{3} = 90 \sqrt{3} - 60$
⇒ $h = 90 - \frac{60}{\sqrt{3}}$ = $90 - 34.64 = 55.36$ m
∴ Height of the first tower = $D E = h = 55.36$ m

Answer:

Let PQ be the chimney and AB be the tower.

We have,

$AB = 40 m, ∠ APB = 30 ° and ∠ PAQ = 60 ° In ∆ ABP, \tan 30 ° = \frac{AB}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{40}{AP} \Rightarrow AP = 40 \sqrt{3} m Now, in ∆ APQ, \tan 60 ° = \frac{PQ}{AP} \Rightarrow \sqrt{3} = \frac{PQ}{40 \sqrt{3}} ∴ PQ = 120 m$

So, the height of the chimney is 120 m.

Hence, the height of the chimney meets the pollution norms.

In this question, management of air pollution has been shown.
a

Question 19:

Let PQ be the chimney and AB be the tower.

We have,

$AB = 40 m, ∠ APB = 30 ° and ∠ PAQ = 60 ° In ∆ ABP, \tan 30 ° = \frac{AB}{AP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{40}{AP} \Rightarrow AP = 40 \sqrt{3} m Now, in ∆ APQ, \tan 60 ° = \frac{PQ}{AP} \Rightarrow \sqrt{3} = \frac{PQ}{40 \sqrt{3}} ∴ PQ = 120 m$

So, the height of the chimney is 120 m.

Hence, the height of the chimney meets the pollution norms.

In this question, management of air pollution has been shown.
a

Answer:

Let AB be the 7-m high building and CD be the cable tower.

We have,

$AB = 7 m, ∠ CAE = 60 °, ∠ DAE = ∠ ADB = 45 ° Also, DE = AB = 7 m In ∆ ABD, \tan 45 ° = \frac{AB}{BD} \Rightarrow 1 = \frac{7}{BD} \Rightarrow BD = 7 m So, AE = BD = 7 m Also, in ∆ ACE, \tan 60 ° = \frac{CE}{AE} \Rightarrow \sqrt{3} = \frac{CE}{7} \Rightarrow CE = 7 \sqrt{3} m Now, CD = CE + DE = 7 \sqrt{3} + 7 = 7 (\sqrt{3} + 1) m = 7 (1.732 + 1) = 7 (2.732) = 19.124 \approx 19.12 m$

So, the height of the tower is 19.12 m.

Question 20:

Let AB be the 7-m high building and CD be the cable tower.

We have,

$AB = 7 m, ∠ CAE = 60 °, ∠ DAE = ∠ ADB = 45 ° Also, DE = AB = 7 m In ∆ ABD, \tan 45 ° = \frac{AB}{BD} \Rightarrow 1 = \frac{7}{BD} \Rightarrow BD = 7 m So, AE = BD = 7 m Also, in ∆ ACE, \tan 60 ° = \frac{CE}{AE} \Rightarrow \sqrt{3} = \frac{CE}{7} \Rightarrow CE = 7 \sqrt{3} m Now, CD = CE + DE = 7 \sqrt{3} + 7 = 7 (\sqrt{3} + 1) m = 7 (1.732 + 1) = 7 (2.732) = 19.124 \approx 19.12 m$

So, the height of the tower is 19.12 m.

Answer:

Let PQ be the tower.

We have,

$AB = 20 m, ∠ PAQ = 30 ° and ∠ PBQ = 60 ° Let BQ = x and PQ = h In ∆ PBQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x \sqrt{3} . . . . . (i) Also, in ∆ APQ, \tan 30 ° = \frac{PQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x \sqrt{3}}{20 + x} [Using (i)] \Rightarrow 20 + x = 3 x \Rightarrow 3 x - x = 20 \Rightarrow 2 x = 20 \Rightarrow x = \frac{20}{2} \Rightarrow x = 10 m From (i), h = 10 \sqrt{3} = 10 \times 1.732 = 17.32 m Also, AQ = AB + BQ = 20 + 10 = 30 m$

So, the height of the tower is 17.32 m and its distance from the point A is 30 m.

Question 21:

Let PQ be the tower.

We have,

$AB = 20 m, ∠ PAQ = 30 ° and ∠ PBQ = 60 ° Let BQ = x and PQ = h In ∆ PBQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x \sqrt{3} . . . . . (i) Also, in ∆ APQ, \tan 30 ° = \frac{PQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x \sqrt{3}}{20 + x} [Using (i)] \Rightarrow 20 + x = 3 x \Rightarrow 3 x - x = 20 \Rightarrow 2 x = 20 \Rightarrow x = \frac{20}{2} \Rightarrow x = 10 m From (i), h = 10 \sqrt{3} = 10 \times 1.732 = 17.32 m Also, AQ = AB + BQ = 20 + 10 = 30 m$

So, the height of the tower is 17.32 m and its distance from the point A is 30 m.

Answer:

Let PQ be the tower.

We have,

$AB = 10 m, ∠ MAP = 30 ° and ∠ PBQ = 60 ° Also, MQ = AB = 10 m Let BQ = x and PQ = h So, AM = BQ = x and PM = PQ - MQ = h - 10 In ∆ BPQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} . . . . . (i) Now, in ∆ AMP, \tan 30 ° = \frac{PM}{AM} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h - 10}{x} \Rightarrow h \sqrt{3} - 10 \sqrt{3} = x \Rightarrow h \sqrt{3} - 10 \sqrt{3} = \frac{h}{\sqrt{3}} [Using (i)] \Rightarrow 3 h - 30 = h \Rightarrow 3 h - h = 30 \Rightarrow 2 h = 30 \Rightarrow h = \frac{30}{2} ∴ h = 15 m$

So, the height of the tower is 15 m.

Question 22:

Let PQ be the tower.

We have,

$AB = 10 m, ∠ MAP = 30 ° and ∠ PBQ = 60 ° Also, MQ = AB = 10 m Let BQ = x and PQ = h So, AM = BQ = x and PM = PQ - MQ = h - 10 In ∆ BPQ, \tan 60 ° = \frac{PQ}{BQ} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} . . . . . (i) Now, in ∆ AMP, \tan 30 ° = \frac{PM}{AM} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h - 10}{x} \Rightarrow h \sqrt{3} - 10 \sqrt{3} = x \Rightarrow h \sqrt{3} - 10 \sqrt{3} = \frac{h}{\sqrt{3}} [Using (i)] \Rightarrow 3 h - 30 = h \Rightarrow 3 h - h = 30 \Rightarrow 2 h = 30 \Rightarrow h = \frac{30}{2} ∴ h = 15 m$

So, the height of the tower is 15 m.

Answer:

Let AD be the tower and BC be the cliff.

We have,

$BC = 60 \sqrt{3} m, ∠ CDE = 45 ° and ∠ BAC = 60 ° Let AD = h \Rightarrow BE = AD = h \Rightarrow CE = BC - BE = 60 \sqrt{3} - h In ∆ CDE, \tan 45 ° = \frac{CE}{DE} \Rightarrow 1 = \frac{60 \sqrt{3} - h}{DE} \Rightarrow DE = 60 \sqrt{3} - h \Rightarrow AB = DE = 60 \sqrt{3} - h . . . . . (i) Now, in ∆ ABC, \tan 60 ° = \frac{BC}{AB} \Rightarrow \sqrt{3} = \frac{60 \sqrt{3}}{60 \sqrt{3} - h} [Using (i)] \Rightarrow 180 - h \sqrt{3} = 60 \sqrt{3} \Rightarrow h \sqrt{3} = 180 - 60 \sqrt{3} \Rightarrow h = \frac{180 - 60 \sqrt{3}}{\sqrt{3}} \Rightarrow h = \frac{180 - 60 \sqrt{3}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow h = \frac{180 \sqrt{3} - 180}{3} \Rightarrow h = \frac{180 (\sqrt{3} - 1)}{3} ∴ h = 60 (\sqrt{3} - 1) = 60 (1.732 - 1) = 60 (0.732) Also, h = 43.92 m$

So, the height of the tower is 43.92 m.

Question 23:

Let AD be the tower and BC be the cliff.

We have,

$BC = 60 \sqrt{3} m, ∠ CDE = 45 ° and ∠ BAC = 60 ° Let AD = h \Rightarrow BE = AD = h \Rightarrow CE = BC - BE = 60 \sqrt{3} - h In ∆ CDE, \tan 45 ° = \frac{CE}{DE} \Rightarrow 1 = \frac{60 \sqrt{3} - h}{DE} \Rightarrow DE = 60 \sqrt{3} - h \Rightarrow AB = DE = 60 \sqrt{3} - h . . . . . (i) Now, in ∆ ABC, \tan 60 ° = \frac{BC}{AB} \Rightarrow \sqrt{3} = \frac{60 \sqrt{3}}{60 \sqrt{3} - h} [Using (i)] \Rightarrow 180 - h \sqrt{3} = 60 \sqrt{3} \Rightarrow h \sqrt{3} = 180 - 60 \sqrt{3} \Rightarrow h = \frac{180 - 60 \sqrt{3}}{\sqrt{3}} \Rightarrow h = \frac{180 - 60 \sqrt{3}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow h = \frac{180 \sqrt{3} - 180}{3} \Rightarrow h = \frac{180 (\sqrt{3} - 1)}{3} ∴ h = 60 (\sqrt{3} - 1) = 60 (1.732 - 1) = 60 (0.732) Also, h = 43.92 m$

So, the height of the tower is 43.92 m.

Answer:

Let $A B$ be the deck of the ship above the water level and $D E$ be the cliff.
Now,
$A B = 16$ m such that $C D = 16$ m and ∠â€‹ $B D A = 30^{o}$ and ∠ $E B C = 60^{o}$ .
If AD = x m and $D E = h$ m, then $C E = (h - 16)$ m.

In the right âˆ† $B A D$ , we have:
$\frac{A B}{A D} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{16}{x} = \frac{1}{\sqrt{3}}$

⇒ $x = 16 \sqrt{3}$ $= 27.68$ m

In the right âˆ† $E B C$ , we have:
$\frac{E C}{B C} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{(h - 16)}{x} = \sqrt{3}$
⇒ $h - 16 = x \sqrt{3}$
⇒ $h - 16 = 16 \sqrt{3} \times \sqrt{3} = 48$ [âˆµ $x = 16 \sqrt{3}$ ]
⇒ $h = 48 + 16 = 64$ m

∴ Distance of the cliff from the deck of the ship = $A D = x = 27.68$ m
And,
Height of the cliff = $D E = h = 64$ m

Question 24:

Let $A B$ be the deck of the ship above the water level and $D E$ be the cliff.
Now,
$A B = 16$ m such that $C D = 16$ m and ∠â€‹ $B D A = 30^{o}$ and ∠ $E B C = 60^{o}$ .
If AD = x m and $D E = h$ m, then $C E = (h - 16)$ m.

In the right âˆ† $B A D$ , we have:
$\frac{A B}{A D} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{16}{x} = \frac{1}{\sqrt{3}}$

⇒ $x = 16 \sqrt{3}$ $= 27.68$ m

In the right âˆ† $E B C$ , we have:
$\frac{E C}{B C} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{(h - 16)}{x} = \sqrt{3}$
⇒ $h - 16 = x \sqrt{3}$
⇒ $h - 16 = 16 \sqrt{3} \times \sqrt{3} = 48$ [âˆµ $x = 16 \sqrt{3}$ ]
⇒ $h = 48 + 16 = 64$ m

∴ Distance of the cliff from the deck of the ship = $A D = x = 27.68$ m
And,
Height of the cliff = $D E = h = 64$ m

Answer:

We have,

$XY = 40 m, ∠ PXQ = 60 ° and ∠ MYQ = 45 ° Let PQ = h Also, MP = XY = 40 m, MQ = PQ - MP = h - 40 In ∆ MYQ, \tan 45 ° = \frac{MQ}{MY} \Rightarrow 1 = \frac{h - 40}{MY} \Rightarrow MY = h - 40 \Rightarrow PX = MY = h - 40 . . . . . (i) Now, in ∆ MXQ, \tan 60 ° = \frac{PQ}{PX} \Rightarrow \sqrt{3} = \frac{h}{h - 40} [From (i)] \Rightarrow h \sqrt{3} - 40 \sqrt{3} = h \Rightarrow h \sqrt{3} - h = 40 \sqrt{3} \Rightarrow h (\sqrt{3} - 1) = 40 \sqrt{3} \Rightarrow h = \frac{40 \sqrt{3}}{(\sqrt{3} - 1)} \Rightarrow h = \frac{40 \sqrt{3}}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \Rightarrow h = \frac{40 \sqrt{3} (\sqrt{3} + 1)}{(3 - 1)} \Rightarrow h = \frac{40 \sqrt{3} (\sqrt{3} + 1)}{2} \Rightarrow h = 20 \sqrt{3} (\sqrt{3} + 1) \Rightarrow h = 60 + 20 \sqrt{3} \Rightarrow h = 60 + 20 \times 1.73 \Rightarrow h = 60 + 34.6 ∴ h = 94.6 m$

So, the height of the tower PQ is 94.6 m.

Question 25:

We have,

$XY = 40 m, ∠ PXQ = 60 ° and ∠ MYQ = 45 ° Let PQ = h Also, MP = XY = 40 m, MQ = PQ - MP = h - 40 In ∆ MYQ, \tan 45 ° = \frac{MQ}{MY} \Rightarrow 1 = \frac{h - 40}{MY} \Rightarrow MY = h - 40 \Rightarrow PX = MY = h - 40 . . . . . (i) Now, in ∆ MXQ, \tan 60 ° = \frac{PQ}{PX} \Rightarrow \sqrt{3} = \frac{h}{h - 40} [From (i)] \Rightarrow h \sqrt{3} - 40 \sqrt{3} = h \Rightarrow h \sqrt{3} - h = 40 \sqrt{3} \Rightarrow h (\sqrt{3} - 1) = 40 \sqrt{3} \Rightarrow h = \frac{40 \sqrt{3}}{(\sqrt{3} - 1)} \Rightarrow h = \frac{40 \sqrt{3}}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \Rightarrow h = \frac{40 \sqrt{3} (\sqrt{3} + 1)}{(3 - 1)} \Rightarrow h = \frac{40 \sqrt{3} (\sqrt{3} + 1)}{2} \Rightarrow h = 20 \sqrt{3} (\sqrt{3} + 1) \Rightarrow h = 60 + 20 \sqrt{3} \Rightarrow h = 60 + 20 \times 1.73 \Rightarrow h = 60 + 34.6 ∴ h = 94.6 m$

So, the height of the tower PQ is 94.6 m.

Answer:

Let the height of flying of the aeroplane be PQ = BC and point A be the point of observation.

We have,

$PQ = BC = 2500 m, ∠ PAQ = 45 ° and ∠ BAC = 30 ° In ∆ PAQ, \tan 45 ° = \frac{PQ}{AQ} \Rightarrow 1 = \frac{2500}{AQ} \Rightarrow AQ = 2500 m Also, in ∆ ABC, \tan 30 ° = \frac{BC}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{2500}{AC} \Rightarrow AC = 2500 \sqrt{3} m Now, QC = AC - AQ = 2500 \sqrt{3} - 2500 = 2500 (\sqrt{3} - 1) m = 2500 (1.732 - 1) = 2500 (0.732) = 1830 m \Rightarrow PB = QC = 1830 m So, the speed of the aeroplane = \frac{PB}{15} = \frac{1830}{15} = 122 m / s = 122 \times \frac{3600}{1000} km / h = 439.2 km / h$

So, the speed of the aeroplane is 122 m/s or 439.2 km/h.

Question 26:

Let the height of flying of the aeroplane be PQ = BC and point A be the point of observation.

We have,

$PQ = BC = 2500 m, ∠ PAQ = 45 ° and ∠ BAC = 30 ° In ∆ PAQ, \tan 45 ° = \frac{PQ}{AQ} \Rightarrow 1 = \frac{2500}{AQ} \Rightarrow AQ = 2500 m Also, in ∆ ABC, \tan 30 ° = \frac{BC}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{2500}{AC} \Rightarrow AC = 2500 \sqrt{3} m Now, QC = AC - AQ = 2500 \sqrt{3} - 2500 = 2500 (\sqrt{3} - 1) m = 2500 (1.732 - 1) = 2500 (0.732) = 1830 m \Rightarrow PB = QC = 1830 m So, the speed of the aeroplane = \frac{PB}{15} = \frac{1830}{15} = 122 m / s = 122 \times \frac{3600}{1000} km / h = 439.2 km / h$

So, the speed of the aeroplane is 122 m/s or 439.2 km/h.

Answer:

Let $A B$ be the tower.
We have:
$C D = 150$ m, ∠ $A C B = 30^{o}$ and ∠ $A D B = 60^{o}$
Let:
$A B = h$ m and $B D = x$ m

In the right âˆ† $A B D$ , we have:
$\frac{A B}{A D} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{h}{x} = \sqrt{3}$
⇒ $x = \frac{h}{\sqrt{3}}$
Now, in the right âˆ† $A C B$ , we have:
$\frac{A B}{B C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x + 150} = \frac{1}{\sqrt{3}}$
⇒ $\sqrt{3} h = x + 150$

On putting $x = \frac{h}{\sqrt{3}}$ in the above equation, we get:
$\sqrt{3} h = \frac{h}{\sqrt{3}} + 150$
⇒ $3 h = h + 150 \sqrt{3}$
⇒ $2 h = 150 \sqrt{3}$
⇒ $h = \frac{150 \sqrt{3}}{2} = 75 \sqrt{3} = 75 \times 1.732 = 129.9$ m
Hence, the height of the tower is 129.9 m.

Question 27:

Let $A B$ be the tower.
We have:
$C D = 150$ m, ∠ $A C B = 30^{o}$ and ∠ $A D B = 60^{o}$
Let:
$A B = h$ m and $B D = x$ m

In the right âˆ† $A B D$ , we have:
$\frac{A B}{A D} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{h}{x} = \sqrt{3}$
⇒ $x = \frac{h}{\sqrt{3}}$
Now, in the right âˆ† $A C B$ , we have:
$\frac{A B}{B C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x + 150} = \frac{1}{\sqrt{3}}$
⇒ $\sqrt{3} h = x + 150$

On putting $x = \frac{h}{\sqrt{3}}$ in the above equation, we get:
$\sqrt{3} h = \frac{h}{\sqrt{3}} + 150$
⇒ $3 h = h + 150 \sqrt{3}$
⇒ $2 h = 150 \sqrt{3}$
⇒ $h = \frac{150 \sqrt{3}}{2} = 75 \sqrt{3} = 75 \times 1.732 = 129.9$ m
Hence, the height of the tower is 129.9 m.

Answer:

Let $O A$ be the lighthouse and B and C be the two positions of the ship.
Thus, we have:
$O A = 100$ m, ∠ $O B A = 30^{o}$ and ∠ $O C A = 60^{o}$

Let:
$O C = x$ m and $B C = y$ m
In the right âˆ† $O A C$ , we have:
$\frac{O A}{O C} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{100}{x} = \sqrt{3}$
⇒ $x = \frac{100}{\sqrt{3}}$ m
Now, in the right âˆ† $O B A$ , we have:
$\frac{O A}{O B} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{100}{x + y} = \frac{1}{\sqrt{3}}$
⇒ $x + y = 100 \sqrt{3}$

On putting $x = \frac{100}{\sqrt{3}}$ in the above equation, we get:
$y = 100 \sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = 115.47$ m

∴ Distance travelled by the ship during the period of observation = $B C = y = 115.47$ m

Question 28:

Let $O A$ be the lighthouse and B and C be the two positions of the ship.
Thus, we have:
$O A = 100$ m, ∠ $O B A = 30^{o}$ and ∠ $O C A = 60^{o}$

Let:
$O C = x$ m and $B C = y$ m
In the right âˆ† $O A C$ , we have:
$\frac{O A}{O C} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{100}{x} = \sqrt{3}$
⇒ $x = \frac{100}{\sqrt{3}}$ m
Now, in the right âˆ† $O B A$ , we have:
$\frac{O A}{O B} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{100}{x + y} = \frac{1}{\sqrt{3}}$
⇒ $x + y = 100 \sqrt{3}$

On putting $x = \frac{100}{\sqrt{3}}$ in the above equation, we get:
$y = 100 \sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = 115.47$ m

∴ Distance travelled by the ship during the period of observation = $B C = y = 115.47$ m

Answer:

Let $A$ and $B$ be two points on the banks on the opposite side of the river and $P$ be the point on the bridge at a height of 2.5 m.
Thus, we have:
$D P = 2.5$ m, ∠PAD $= 30^{o}$ and ∠ $P B D = 45^{o}$
In the right âˆ† $A P D$ , we have:
$\frac{D P}{A D} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{2.5}{A D} = \frac{1}{\sqrt{3}}$
⇒ $A D = 2.5 \sqrt{3}$ m
In the right âˆ† $P D B$ , we have:
$\frac{D P}{B D} = \tan 45^{o} = 1$
⇒ $\frac{2.5}{B D} = 1$
⇒ $B D = 2.5$ m
â€‹
∴ Width of the river = $A B = (A D + B D) = (2.5 \sqrt{3} + 2.5) = 6.83$ m

Question 29:

Let $A$ and $B$ be two points on the banks on the opposite side of the river and $P$ be the point on the bridge at a height of 2.5 m.
Thus, we have:
$D P = 2.5$ m, ∠PAD $= 30^{o}$ and ∠ $P B D = 45^{o}$
In the right âˆ† $A P D$ , we have:
$\frac{D P}{A D} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{2.5}{A D} = \frac{1}{\sqrt{3}}$
⇒ $A D = 2.5 \sqrt{3}$ m
In the right âˆ† $P D B$ , we have:
$\frac{D P}{B D} = \tan 45^{o} = 1$
⇒ $\frac{2.5}{B D} = 1$
⇒ $B D = 2.5$ m
â€‹
∴ Width of the river = $A B = (A D + B D) = (2.5 \sqrt{3} + 2.5) = 6.83$ m

Answer:

Let $A B$ be the tower and $C and D$ be two points such that $A C = 4$ m and $A D = 9$ m.
Let:
$A B = h$ m, ∠ $B C A = θ$ and ∠ $B D A = 90^{°} - θ$

In the right âˆ†BCA, we have:
$\tan θ = \frac{A B}{A C} \Rightarrow \tan θ = \frac{h}{4} . . . (1)$
In the right âˆ†BDA, we have:
$\tan (90 ° - θ) = \frac{A B}{A D} \Rightarrow \cot θ = \frac{h}{9} [\tan (90 ° - θ) = \cot θ] \Rightarrow \frac{1}{\tan θ} = \frac{h}{9} . . . (2) [\cot θ = \frac{1}{\tan θ}]$
Multiplying equations (1) and (2), we get:
$\tan θ \times \frac{1}{\tan θ} = \frac{h}{4} \times \frac{h}{9} \Rightarrow 1 = \frac{h^{2}}{36}$
⇒ 36 = h²
⇒ h = ±6

Height of a tower cannot be negative.
∴ Height of the tower = 6 m

Question 30:

Let $A B$ be the tower and $C and D$ be two points such that $A C = 4$ m and $A D = 9$ m.
Let:
$A B = h$ m, ∠ $B C A = θ$ and ∠ $B D A = 90^{°} - θ$

In the right âˆ†BCA, we have:
$\tan θ = \frac{A B}{A C} \Rightarrow \tan θ = \frac{h}{4} . . . (1)$
In the right âˆ†BDA, we have:
$\tan (90 ° - θ) = \frac{A B}{A D} \Rightarrow \cot θ = \frac{h}{9} [\tan (90 ° - θ) = \cot θ] \Rightarrow \frac{1}{\tan θ} = \frac{h}{9} . . . (2) [\cot θ = \frac{1}{\tan θ}]$
Multiplying equations (1) and (2), we get:
$\tan θ \times \frac{1}{\tan θ} = \frac{h}{4} \times \frac{h}{9} \Rightarrow 1 = \frac{h^{2}}{36}$
⇒ 36 = h²
⇒ h = ±6

Height of a tower cannot be negative.
∴ Height of the tower = 6 m

Answer:

Let AB and CD be the two opposite walls of the room and the foot of the ladder be fixed at the point O on the ground.

We have,

$AO = CO = 6 m, ∠ AOB = 60 ° and ∠ COD = 45 ° In ∆ ABO, \cos 60 ° = \frac{BO}{AO} \Rightarrow \frac{1}{2} = \frac{BO}{6} \Rightarrow BO = \frac{6}{2} \Rightarrow BO = 3 m Also, in ∆ CDO, \cos 45 ° = \frac{DO}{CO} \Rightarrow \frac{1}{\sqrt{2}} = \frac{DO}{6} \Rightarrow DO = \frac{6}{\sqrt{2}} \Rightarrow DO = \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \Rightarrow DO = \frac{6 \sqrt{2}}{2} \Rightarrow DO = 3 \sqrt{2} m Now, the distance between two walls of the room = BD = BO + DO = 3 + 3 \sqrt{2} = 3 (1 + \sqrt{2}) = 3 (1 + 1.414) = 3 (2.414) = 7.242 \approx 7.24 m$

So, the distance between two walls of the room is 7.24 m.

Question 31:

Let AB and CD be the two opposite walls of the room and the foot of the ladder be fixed at the point O on the ground.

We have,

$AO = CO = 6 m, ∠ AOB = 60 ° and ∠ COD = 45 ° In ∆ ABO, \cos 60 ° = \frac{BO}{AO} \Rightarrow \frac{1}{2} = \frac{BO}{6} \Rightarrow BO = \frac{6}{2} \Rightarrow BO = 3 m Also, in ∆ CDO, \cos 45 ° = \frac{DO}{CO} \Rightarrow \frac{1}{\sqrt{2}} = \frac{DO}{6} \Rightarrow DO = \frac{6}{\sqrt{2}} \Rightarrow DO = \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \Rightarrow DO = \frac{6 \sqrt{2}}{2} \Rightarrow DO = 3 \sqrt{2} m Now, the distance between two walls of the room = BD = BO + DO = 3 + 3 \sqrt{2} = 3 (1 + \sqrt{2}) = 3 (1 + 1.414) = 3 (2.414) = 7.242 \approx 7.24 m$

So, the distance between two walls of the room is 7.24 m.

Answer:

Let OP be the tower and points A and B be the positions of the cars.

We have,

$AB = 100 m, ∠ OAP = 60 ° and ∠ OBP = 45 ° Let OP = h In ∆ AOP, \tan 60 ° = \frac{OP}{OA} \Rightarrow \sqrt{3} = \frac{h}{OA} \Rightarrow OA = \frac{h}{\sqrt{3}} Also, in ∆ BOP, \tan 45 ° = \frac{OP}{OB} \Rightarrow 1 = \frac{h}{OB} \Rightarrow OB = h Now, OB - OA = 100 \Rightarrow h - \frac{h}{\sqrt{3}} = 100 \Rightarrow \frac{h \sqrt{3} - h}{\sqrt{3}} = 100 \Rightarrow \frac{h (\sqrt{3} - 1)}{\sqrt{3}} = 100 \Rightarrow h = \frac{100 \sqrt{3}}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \Rightarrow h = \frac{100 \sqrt{3} (\sqrt{3} + 1)}{(3 - 1)} \Rightarrow h = \frac{100 (3 + \sqrt{3})}{2} \Rightarrow h = 50 (3 + 1.732) \Rightarrow h = 50 (4.732) ∴ h = 236.6 m$

So, the height of the tower is 236.6 m.

Disclaimer: The answer given in the texbook is incorrect. The same has been rectified above.

Question 32:

Let OP be the tower and points A and B be the positions of the cars.

We have,

$AB = 100 m, ∠ OAP = 60 ° and ∠ OBP = 45 ° Let OP = h In ∆ AOP, \tan 60 ° = \frac{OP}{OA} \Rightarrow \sqrt{3} = \frac{h}{OA} \Rightarrow OA = \frac{h}{\sqrt{3}} Also, in ∆ BOP, \tan 45 ° = \frac{OP}{OB} \Rightarrow 1 = \frac{h}{OB} \Rightarrow OB = h Now, OB - OA = 100 \Rightarrow h - \frac{h}{\sqrt{3}} = 100 \Rightarrow \frac{h \sqrt{3} - h}{\sqrt{3}} = 100 \Rightarrow \frac{h (\sqrt{3} - 1)}{\sqrt{3}} = 100 \Rightarrow h = \frac{100 \sqrt{3}}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \Rightarrow h = \frac{100 \sqrt{3} (\sqrt{3} + 1)}{(3 - 1)} \Rightarrow h = \frac{100 (3 + \sqrt{3})}{2} \Rightarrow h = 50 (3 + 1.732) \Rightarrow h = 50 (4.732) ∴ h = 236.6 m$

So, the height of the tower is 236.6 m.

Disclaimer: The answer given in the texbook is incorrect. The same has been rectified above.

Answer:

Let AC be the pole and BD be the ladder.

We have,

$AC = 4 m, AB = 1 m and ∠ BDC = 60 ° And, BC = AC - AB = 4 - 1 = 3 m In ∆ BDC, \sin 60 ° = \frac{BC}{BD} \Rightarrow \frac{\sqrt{3}}{2} = \frac{3}{BD} \Rightarrow BD = \frac{3 \times 2}{\sqrt{3}} \Rightarrow BD = 2 \sqrt{3} \Rightarrow BD = 2 \times 1.73 ∴ BD = 3.46 m$

So, he should use 3.46 m long ladder to reach the required position.

Question 33:

Let AC be the pole and BD be the ladder.

We have,

$AC = 4 m, AB = 1 m and ∠ BDC = 60 ° And, BC = AC - AB = 4 - 1 = 3 m In ∆ BDC, \sin 60 ° = \frac{BC}{BD} \Rightarrow \frac{\sqrt{3}}{2} = \frac{3}{BD} \Rightarrow BD = \frac{3 \times 2}{\sqrt{3}} \Rightarrow BD = 2 \sqrt{3} \Rightarrow BD = 2 \times 1.73 ∴ BD = 3.46 m$

So, he should use 3.46 m long ladder to reach the required position.

Answer:

We have,

$AB = 60 m, ∠ ACE = 30 ° and ∠ ADB = 60 ° Let BD = CE = x and CD = BE = y \Rightarrow AE = AB - BE = 60 - y In ∆ ACE, \tan 30 ° = \frac{AE}{CE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{60 - y}{x} \Rightarrow x = 60 \sqrt{3} - y \sqrt{3} . . . . . (i) Also, in ∆ ABD, \tan 60 ° = \frac{AB}{BD} \Rightarrow \sqrt{3} = \frac{60}{x} \Rightarrow x = \frac{60}{\sqrt{3}} \Rightarrow x = \frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow x = \frac{60 \sqrt{3}}{3} \Rightarrow x = 20 \sqrt{3} Substituting x = 20 \sqrt{3} in (i), we get 20 \sqrt{3} = 60 \sqrt{3} - y \sqrt{3} \Rightarrow y \sqrt{3} = 60 \sqrt{3} - 20 \sqrt{3} \Rightarrow y \sqrt{3} = 40 \sqrt{3} \Rightarrow y = \frac{40 \sqrt{3}}{\sqrt{3}} \Rightarrow y = 40 m (i) the horizontal distance between AB and CD = BD = x = 20 \sqrt{3} = 20 \times 1.732 = 34.64 m (ii) the height of the lamp post = CD = y = 40 m (iii) the difference between the heights of the building and the lamp post = AB - CD = 60 - 40 = 20 m$

Question 34:

We have,

$AB = 60 m, ∠ ACE = 30 ° and ∠ ADB = 60 ° Let BD = CE = x and CD = BE = y \Rightarrow AE = AB - BE = 60 - y In ∆ ACE, \tan 30 ° = \frac{AE}{CE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{60 - y}{x} \Rightarrow x = 60 \sqrt{3} - y \sqrt{3} . . . . . (i) Also, in ∆ ABD, \tan 60 ° = \frac{AB}{BD} \Rightarrow \sqrt{3} = \frac{60}{x} \Rightarrow x = \frac{60}{\sqrt{3}} \Rightarrow x = \frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow x = \frac{60 \sqrt{3}}{3} \Rightarrow x = 20 \sqrt{3} Substituting x = 20 \sqrt{3} in (i), we get 20 \sqrt{3} = 60 \sqrt{3} - y \sqrt{3} \Rightarrow y \sqrt{3} = 60 \sqrt{3} - 20 \sqrt{3} \Rightarrow y \sqrt{3} = 40 \sqrt{3} \Rightarrow y = \frac{40 \sqrt{3}}{\sqrt{3}} \Rightarrow y = 40 m (i) the horizontal distance between AB and CD = BD = x = 20 \sqrt{3} = 20 \times 1.732 = 34.64 m (ii) the height of the lamp post = CD = y = 40 m (iii) the difference between the heights of the building and the lamp post = AB - CD = 60 - 40 = 20 m$

Answer:

Suppose AB be the tower of height h meters. Let C be the initial position of the car and let after 12 minutes the car be at D. It is given that the angles of depression at C and D are 30º and 45º respectively.
Let the speed of the car be v meter per minute. Then,
CD = distance travelled by the car in 12 minutes
CD = 12v meters

Suppose the car takes t minutes to reach the tower AB from D. Then DA = vt meters.
$In ∆ DAB, we have \tan 45 ° = \frac{AB}{AD} \Rightarrow 1 = \frac{h}{v t} \Rightarrow h = v t . . . . . (i) In ∆ CAB, we have \tan 30 ° = \frac{AB}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{v t + 12 v} \Rightarrow \sqrt{3} h = v t + 12 v . . . . . (i i)$
Substituting the value of h from equation (i) in equation (ii), we get
$\Rightarrow \sqrt{3} t = t + 12 \Rightarrow t = \frac{12}{\sqrt{3} - 1} = \frac{12}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} = 6 (\sqrt{3} + 1) \min$

Question 35:

Suppose AB be the tower of height h meters. Let C be the initial position of the car and let after 12 minutes the car be at D. It is given that the angles of depression at C and D are 30º and 45º respectively.
Let the speed of the car be v meter per minute. Then,
CD = distance travelled by the car in 12 minutes
CD = 12v meters

Suppose the car takes t minutes to reach the tower AB from D. Then DA = vt meters.
$In ∆ DAB, we have \tan 45 ° = \frac{AB}{AD} \Rightarrow 1 = \frac{h}{v t} \Rightarrow h = v t . . . . . (i) In ∆ CAB, we have \tan 30 ° = \frac{AB}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{v t + 12 v} \Rightarrow \sqrt{3} h = v t + 12 v . . . . . (i i)$
Substituting the value of h from equation (i) in equation (ii), we get
$\Rightarrow \sqrt{3} t = t + 12 \Rightarrow t = \frac{12}{\sqrt{3} - 1} = \frac{12}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} = 6 (\sqrt{3} + 1) \min$

Answer:

Let CD be the height of the aeroplane above the river at some instant. Suppose A and B be two points on both banks of the river in opposite directions.

Height of the aeroplane above the river, CD = 300 m
Now,
$∠$ CAD = $∠$ ADX = 60º (Alternate angles)
$∠$ CBD = $∠$ BDY = 45º (Alternate angles)
In right âˆ†ACD,
$\tan 60 ° = \frac{CD}{AC} \Rightarrow \sqrt{3} = \frac{300}{AC} \Rightarrow AC = \frac{300}{\sqrt{3}} = 100 \sqrt{3} m$
In right âˆ†BCD,
$\tan 45 ° = \frac{CD}{BC} \Rightarrow 1 = \frac{300}{BC} \Rightarrow BC = 300 m$
∴ Width of the river, AB
= BC + AC
$= 300 + 100 \sqrt{3} = 300 + 100 \times 1.73 = 300 + 173 = 473 m$
Thus, the width of the river is 473 m.

Question 36:

Let CD be the height of the aeroplane above the river at some instant. Suppose A and B be two points on both banks of the river in opposite directions.

Height of the aeroplane above the river, CD = 300 m
Now,
$∠$ CAD = $∠$ ADX = 60º (Alternate angles)
$∠$ CBD = $∠$ BDY = 45º (Alternate angles)
In right âˆ†ACD,
$\tan 60 ° = \frac{CD}{AC} \Rightarrow \sqrt{3} = \frac{300}{AC} \Rightarrow AC = \frac{300}{\sqrt{3}} = 100 \sqrt{3} m$
In right âˆ†BCD,
$\tan 45 ° = \frac{CD}{BC} \Rightarrow 1 = \frac{300}{BC} \Rightarrow BC = 300 m$
∴ Width of the river, AB
= BC + AC
$= 300 + 100 \sqrt{3} = 300 + 100 \times 1.73 = 300 + 173 = 473 m$
Thus, the width of the river is 473 m.

Answer:

Let BC be the 20 m high building and AB be the communication tower of height h fixed on top of the building. Let D be a point on ground such that CD = x m and angles of elevation made from this point to top and bottom of tower are $45 °$ and $60 °$ .
In $△ BCD : \tan 45 ° = \frac{BC}{CD} = \frac{20}{x}$
$\Rightarrow 1 = \frac{20}{x} \Rightarrow x = 20 m .$

Also, in $△ ACD : \tan 60 ° = \frac{AC}{CD} = \frac{20 + h}{x}$
$\Rightarrow \sqrt{3} = \frac{20 + h}{x} \Rightarrow \sqrt{3} = \frac{20 + h}{20} \Rightarrow 20 + h = 20 \sqrt{3} \Rightarrow h = 20 \sqrt{3} - 20 \Rightarrow h = 20 (\sqrt{3} - 1) = 20 (1.73 - 1) = 20 \times 0.73 \Rightarrow h = 14.64 m .$

Question 37:

Let BC be the 20 m high building and AB be the communication tower of height h fixed on top of the building. Let D be a point on ground such that CD = x m and angles of elevation made from this point to top and bottom of tower are $45 °$ and $60 °$ .
In $△ BCD : \tan 45 ° = \frac{BC}{CD} = \frac{20}{x}$
$\Rightarrow 1 = \frac{20}{x} \Rightarrow x = 20 m .$

Also, in $△ ACD : \tan 60 ° = \frac{AC}{CD} = \frac{20 + h}{x}$
$\Rightarrow \sqrt{3} = \frac{20 + h}{x} \Rightarrow \sqrt{3} = \frac{20 + h}{20} \Rightarrow 20 + h = 20 \sqrt{3} \Rightarrow h = 20 \sqrt{3} - 20 \Rightarrow h = 20 (\sqrt{3} - 1) = 20 (1.73 - 1) = 20 \times 0.73 \Rightarrow h = 14.64 m .$

Answer:

Let PQ be the hill of height h km. Let R and S be two consecutive kilometre stones, so the distance between them is 1 km.
Let QR = x km.

$In ∆ PQR, \tan 45 ° = \frac{PQ}{QR} \Rightarrow 1 = \frac{h}{x} \Rightarrow h = x . . . (i)$

$In ∆ PQS, \tan 30 ° = \frac{PQ}{QS} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 1} \Rightarrow \sqrt{3} h = x + 1 . . . (ii)$
From equation (i) and (ii) we get,
$\sqrt{3} h = h + 1 \Rightarrow h (\sqrt{3} - 1) = 1 \Rightarrow h = \frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{(\sqrt{3} - 1) (\sqrt{3} + 1)} \Rightarrow h = \frac{\sqrt{3} + 1}{2} = \frac{2.73}{2} = 1.365 km$
Hence, the height of the hill is 1.365 km.

Question 38:

Let PQ be the hill of height h km. Let R and S be two consecutive kilometre stones, so the distance between them is 1 km.
Let QR = x km.

$In ∆ PQR, \tan 45 ° = \frac{PQ}{QR} \Rightarrow 1 = \frac{h}{x} \Rightarrow h = x . . . (i)$

$In ∆ PQS, \tan 30 ° = \frac{PQ}{QS} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 1} \Rightarrow \sqrt{3} h = x + 1 . . . (ii)$
From equation (i) and (ii) we get,
$\sqrt{3} h = h + 1 \Rightarrow h (\sqrt{3} - 1) = 1 \Rightarrow h = \frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3} + 1}{(\sqrt{3} - 1) (\sqrt{3} + 1)} \Rightarrow h = \frac{\sqrt{3} + 1}{2} = \frac{2.73}{2} = 1.365 km$
Hence, the height of the hill is 1.365 km.

Answer:

Let AB be the vertically standing pole of height h units and CB be the length of its shadow of s units.
Since, the ratio of length of pole and its shadow at some time of day is given to be $\sqrt{3} : 1 .$
$\Rightarrow \frac{AB}{BC} = \frac{\sqrt{3}}{1} .$
$∴$ In $△ ABC :$
$\tan (θ) = \frac{AB}{BC} = \frac{\sqrt{3}}{1} = \sqrt{3} \Rightarrow \tan (θ) = \tan 60 ° \Rightarrow θ = 60 ° .$

Question 39:

Let AB be the vertically standing pole of height h units and CB be the length of its shadow of s units.
Since, the ratio of length of pole and its shadow at some time of day is given to be $\sqrt{3} : 1 .$
$\Rightarrow \frac{AB}{BC} = \frac{\sqrt{3}}{1} .$
$∴$ In $△ ABC :$
$\tan (θ) = \frac{AB}{BC} = \frac{\sqrt{3}}{1} = \sqrt{3} \Rightarrow \tan (θ) = \tan 60 ° \Rightarrow θ = 60 ° .$

Answer:

Let AB be the light house. Suppose C and D be the two positions of the boat.

Here, AB = 100 m.

Let the speed of the boat be v m/min.

So,

CD = v m/min × 2 min = 2v m [Distance = Speed × Time]

In right âˆ†ABC,

$\tan 60 ° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{100 m}{BC} \Rightarrow BC = \frac{100}{\sqrt{3}} = \frac{100 \sqrt{3}}{3} m . . . . . (1)$

In right âˆ†ABD,

$\tan 30 ° = \frac{AB}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100 m}{BC + CD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{\frac{100 \sqrt{3}}{3} + 2 v} [From (1)]$
$\Rightarrow 2 v + \frac{100 \sqrt{3}}{3} = 100 \sqrt{3} \Rightarrow 2 v = 100 \sqrt{3} - \frac{100 \sqrt{3}}{3} \Rightarrow 2 v = 100 \sqrt{3} (1 - \frac{1}{3}) \Rightarrow 2 v = 100 \sqrt{3} \times \frac{2}{3}$
$\Rightarrow v = \frac{100 \times 1.732}{3} = 57.73 m / \min$

Thus, the speed boat is 57.73 m/min.

Question 40:

Let AB be the light house. Suppose C and D be the two positions of the boat.

Here, AB = 100 m.

Let the speed of the boat be v m/min.

So,

CD = v m/min × 2 min = 2v m [Distance = Speed × Time]

In right âˆ†ABC,

$\tan 60 ° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{100 m}{BC} \Rightarrow BC = \frac{100}{\sqrt{3}} = \frac{100 \sqrt{3}}{3} m . . . . . (1)$

In right âˆ†ABD,

$\tan 30 ° = \frac{AB}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100 m}{BC + CD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{100}{\frac{100 \sqrt{3}}{3} + 2 v} [From (1)]$
$\Rightarrow 2 v + \frac{100 \sqrt{3}}{3} = 100 \sqrt{3} \Rightarrow 2 v = 100 \sqrt{3} - \frac{100 \sqrt{3}}{3} \Rightarrow 2 v = 100 \sqrt{3} (1 - \frac{1}{3}) \Rightarrow 2 v = 100 \sqrt{3} \times \frac{2}{3}$
$\Rightarrow v = \frac{100 \times 1.732}{3} = 57.73 m / \min$

Thus, the speed boat is 57.73 m/min.

Answer:

Let AB be the building and CD be the tower.

Draw AE ⊥ CD.

Suppose the height of the tower be h m.

Here, AB = 8 m

∠DAE = ∠ADX = 30º        (Alternate angles)

∠DBC = ∠BDX = 45º        (Alternate angles)

CE = AB = 8 m

∴ DE = CD − CE = (h − 8) m

Distance between the building and tower = BC

In right âˆ†BCD,

$\tan 45 ° = \frac{CD}{BC} \Rightarrow 1 = \frac{h}{BC} \Rightarrow BC = h m$

In right âˆ†AED,

$\tan 30 ° = \frac{DE}{AE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h - 8}{h} (AE = BC) \Rightarrow \sqrt{3} h - 8 \sqrt{3} = h$
$\Rightarrow \sqrt{3} h - h = 8 \sqrt{3} \Rightarrow (\sqrt{3} - 1) h = 8 \sqrt{3} \Rightarrow h = \frac{8 \sqrt{3}}{\sqrt{3} - 1} \Rightarrow h = \frac{8 \sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3} - 1) \times (\sqrt{3} + 1)}$
$\Rightarrow h = \frac{8 \sqrt{3} (\sqrt{3} + 1)}{2} [(a - b) (a + b) = a^{2} - b^{2}] \Rightarrow h = 4 \sqrt{3} (\sqrt{3} + 1) \Rightarrow h = (12 + 4 \sqrt{3}) m$

∴ Height of the tower = $(12 + 4 \sqrt{3}) m$

Also,

Distance between the tower and building = BC = $(12 + 4 \sqrt{3}) m$                  (BC = h m)

Thus, the the height of the tower is $(12 + 4 \sqrt{3}) m$ and the distance between the tower and the building is $(12 + 4 \sqrt{3}) m$ .

Question 1:

Let AB be the building and CD be the tower.

Draw AE ⊥ CD.

Suppose the height of the tower be h m.

Here, AB = 8 m

∠DAE = ∠ADX = 30º        (Alternate angles)

∠DBC = ∠BDX = 45º        (Alternate angles)

CE = AB = 8 m

∴ DE = CD − CE = (h − 8) m

Distance between the building and tower = BC

In right âˆ†BCD,

$\tan 45 ° = \frac{CD}{BC} \Rightarrow 1 = \frac{h}{BC} \Rightarrow BC = h m$

In right âˆ†AED,

$\tan 30 ° = \frac{DE}{AE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h - 8}{h} (AE = BC) \Rightarrow \sqrt{3} h - 8 \sqrt{3} = h$
$\Rightarrow \sqrt{3} h - h = 8 \sqrt{3} \Rightarrow (\sqrt{3} - 1) h = 8 \sqrt{3} \Rightarrow h = \frac{8 \sqrt{3}}{\sqrt{3} - 1} \Rightarrow h = \frac{8 \sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3} - 1) \times (\sqrt{3} + 1)}$
$\Rightarrow h = \frac{8 \sqrt{3} (\sqrt{3} + 1)}{2} [(a - b) (a + b) = a^{2} - b^{2}] \Rightarrow h = 4 \sqrt{3} (\sqrt{3} + 1) \Rightarrow h = (12 + 4 \sqrt{3}) m$

∴ Height of the tower = $(12 + 4 \sqrt{3}) m$

Also,

Distance between the tower and building = BC = $(12 + 4 \sqrt{3}) m$                  (BC = h m)

Thus, the the height of the tower is $(12 + 4 \sqrt{3}) m$ and the distance between the tower and the building is $(12 + 4 \sqrt{3}) m$ .

Answer:

Let AB represents the vertical pole and BC represents the shadow on the ground and θ represents angle of elevation the sun.

$In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{x}{x} (As, the height of the pole, AB = the length of the shadow, BC = x) \Rightarrow \tan θ = 1 \Rightarrow \tan θ = \tan 45 ° ∴ θ = 45 °$

Hence, the correct answer is option (c).

Question 2:

Let AB represents the vertical pole and BC represents the shadow on the ground and θ represents angle of elevation the sun.

$In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{x}{x} (As, the height of the pole, AB = the length of the shadow, BC = x) \Rightarrow \tan θ = 1 \Rightarrow \tan θ = \tan 45 ° ∴ θ = 45 °$

Hence, the correct answer is option (c).

Answer:

Here, AO be the pole; BO be its shadow and $θ$ be the angle of elevation of the sun.

$Let BO = x Then, AO = x \sqrt{3} In ∆ AOB, \tan θ = \frac{AO}{BO} \Rightarrow \tan θ = \frac{x \sqrt{3}}{x} \Rightarrow \tan θ = \sqrt{3} \Rightarrow \tan θ = \tan 60 ° ∴ θ = 60 °$

Hence, the correct answer is option (c).

Question 3:

Here, AO be the pole; BO be its shadow and $θ$ be the angle of elevation of the sun.

$Let BO = x Then, AO = x \sqrt{3} In ∆ AOB, \tan θ = \frac{AO}{BO} \Rightarrow \tan θ = \frac{x \sqrt{3}}{x} \Rightarrow \tan θ = \sqrt{3} \Rightarrow \tan θ = \tan 60 ° ∴ θ = 60 °$

Hence, the correct answer is option (c).

Answer:

(b) 30°

Let $A B$ be the pole and $B C$ be its shadow.

Let $A B = h$ and $B C = x$ such that $x = \sqrt{3} h$ (given) and $θ$ be the angle of elevation.
From âˆ† $A B C$ , we have:
$\frac{A B}{B C} = \tan θ$

⇒ $\frac{h}{x} = \frac{h}{\sqrt{3} h} = \tan θ$
⇒ $\tan θ = \frac{1}{\sqrt{3}}$
⇒ $θ = 30^{o}$

Hence, the angle of elevation is $30^{o}$ .

Question 4:

(b) 30°

Let $A B$ be the pole and $B C$ be its shadow.

Let $A B = h$ and $B C = x$ such that $x = \sqrt{3} h$ (given) and $θ$ be the angle of elevation.
From âˆ† $A B C$ , we have:
$\frac{A B}{B C} = \tan θ$

⇒ $\frac{h}{x} = \frac{h}{\sqrt{3} h} = \tan θ$
⇒ $\tan θ = \frac{1}{\sqrt{3}}$
⇒ $θ = 30^{o}$

Hence, the angle of elevation is $30^{o}$ .

Answer:

Let AB be the pole, BC be its shadow and $θ$ be the sun's elevation.

$We have, AB = 12 m and BC = 4 \sqrt{3} m In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{12}{4 \sqrt{3}} \Rightarrow \tan θ = \frac{3}{\sqrt{3}} \Rightarrow \tan θ = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow \tan θ = \frac{3 \sqrt{3}}{3} \Rightarrow \tan θ = \sqrt{3} \Rightarrow \tan θ = \tan 60 ° ∴ θ = 60 °$

Hence, the correct answer is option (a).

Question 5:

Let AB be the pole, BC be its shadow and $θ$ be the sun's elevation.

$We have, AB = 12 m and BC = 4 \sqrt{3} m In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{12}{4 \sqrt{3}} \Rightarrow \tan θ = \frac{3}{\sqrt{3}} \Rightarrow \tan θ = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow \tan θ = \frac{3 \sqrt{3}}{3} \Rightarrow \tan θ = \sqrt{3} \Rightarrow \tan θ = \tan 60 ° ∴ θ = 60 °$

Hence, the correct answer is option (a).

Answer:

Let AB be a stick and BC be its shadow; and PQ be the tree and QR be its shadow.

We have,

$AB = 5 m, BC = 2 m, PQ = 12.5 m In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{5}{2} . . . . . (i) Now, in ∆ PQR, \tan θ = \frac{PQ}{QR} \Rightarrow \frac{5}{2} = \frac{12.5}{QR} [Using (i)] \Rightarrow QR = \frac{12.5 \times 2}{5} = \frac{25}{5} ∴ QR = 5 m$

Hence, the correct answer is option (d).

Question 6:

Let AB be a stick and BC be its shadow; and PQ be the tree and QR be its shadow.

We have,

$AB = 5 m, BC = 2 m, PQ = 12.5 m In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{5}{2} . . . . . (i) Now, in ∆ PQR, \tan θ = \frac{PQ}{QR} \Rightarrow \frac{5}{2} = \frac{12.5}{QR} [Using (i)] \Rightarrow QR = \frac{12.5 \times 2}{5} = \frac{25}{5} ∴ QR = 5 m$

Hence, the correct answer is option (d).

Answer:

Let AB be the wall and AC be the ladder.

We have,

BC = 2 m and ∠ ACB = 60 ° In ∆ ABC, \cos 60 ° = \frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{2}{AC} ∴ AC = 4 m

Hence, the correct answer is option (d).

Question 7:

Let AB be the wall and AC be the ladder.

We have,

BC = 2 m and ∠ ACB = 60 ° In ∆ ABC, \cos 60 ° = \frac{BC}{AC} \Rightarrow \frac{1}{2} = \frac{2}{AC} ∴ AC = 4 m

Hence, the correct answer is option (d).

Answer:

Let AB be the wall and AC be the ladder.

We have,

$AC = 15 m and ∠ BAC = 60 ° In ∆ ABC, \cos 60 ° = \frac{AB}{AC} \Rightarrow \frac{1}{2} = \frac{AB}{15} ∴ AB = \frac{15}{2} m$

Hence, the correct answer is option (c).

Question 8:

Let AB be the wall and AC be the ladder.

We have,

$AC = 15 m and ∠ BAC = 60 ° In ∆ ABC, \cos 60 ° = \frac{AB}{AC} \Rightarrow \frac{1}{2} = \frac{AB}{15} ∴ AB = \frac{15}{2} m$

Hence, the correct answer is option (c).

Answer:

Let AB be the tower and point C be the point of observation on the ground.

We have,

$BC = 30 m and ∠ ACB = 30 ° In ∆ ABC, \tan 30 ° = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{30} \Rightarrow AB = \frac{30}{\sqrt{3}} \Rightarrow AB = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow AB = \frac{30 \sqrt{3}}{3} ∴ AB = 10 \sqrt{3} m$

Hence, the correct answer is option (b).

Question 9:

Let AB be the tower and point C be the point of observation on the ground.

We have,

$BC = 30 m and ∠ ACB = 30 ° In ∆ ABC, \tan 30 ° = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{30} \Rightarrow AB = \frac{30}{\sqrt{3}} \Rightarrow AB = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \Rightarrow AB = \frac{30 \sqrt{3}}{3} ∴ AB = 10 \sqrt{3} m$

Hence, the correct answer is option (b).

Answer:

Let AB be the tower and point C be the position of the car.

We have,

$AB = 150 m and ∠ ACB = 30 ° In ∆ ABC, \tan 30 ° = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{BC} ∴ BC = 150 \sqrt{3} m$

Hence, the correct answer is option (b).

Question 10:

Let AB be the tower and point C be the position of the car.

We have,

$AB = 150 m and ∠ ACB = 30 ° In ∆ ABC, \tan 30 ° = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{150}{BC} ∴ BC = 150 \sqrt{3} m$

Hence, the correct answer is option (b).

Answer:

Let point A be the position of the kite and AC be its string.

We have,

$AB = 30 m and AC = 60 m Let ∠ ACB = θ In ∆ ABC, \sin θ = \frac{AB}{AC} \Rightarrow \sin θ = \frac{30}{60} \Rightarrow \sin θ = \frac{1}{2} \Rightarrow \sin θ = \sin 30 ° ∴ θ = 30 °$

Hence, the correct answer is option (b).

Question 11:

Let point A be the position of the kite and AC be its string.

We have,

$AB = 30 m and AC = 60 m Let ∠ ACB = θ In ∆ ABC, \sin θ = \frac{AB}{AC} \Rightarrow \sin θ = \frac{30}{60} \Rightarrow \sin θ = \frac{1}{2} \Rightarrow \sin θ = \sin 30 ° ∴ θ = 30 °$

Hence, the correct answer is option (b).

Answer:

Let AB be the cliff and CD be the tower.

We have,

$AB = 20 m Also, CE = AB = 20 m Let ∠ ACB = ∠ CAE = ∠ DAE = θ In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{20}{BC} \Rightarrow \tan θ = \frac{20}{AE} (As, BC = AE) \Rightarrow AE = \frac{20}{\tan θ} . . . . . (i) Also, in ∆ ADE, \tan θ = \frac{DE}{AE} \Rightarrow \tan θ = \frac{DE}{(\frac{20}{\tan θ})} [Using (i)] \Rightarrow \tan θ = \frac{DE \times \tan θ}{20} \Rightarrow DE = \frac{20 \times \tan θ}{\tan θ} \Rightarrow DE = 20 m Now, CD = DE + CE = 20 + 20 ∴ CD = 40 m$

Hence, the correct answer is option (b).

Disclaimer: The answer given in the textbook is incorrect. The same has been rectified above.

Question 12:

Let AB be the cliff and CD be the tower.

We have,

$AB = 20 m Also, CE = AB = 20 m Let ∠ ACB = ∠ CAE = ∠ DAE = θ In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{20}{BC} \Rightarrow \tan θ = \frac{20}{AE} (As, BC = AE) \Rightarrow AE = \frac{20}{\tan θ} . . . . . (i) Also, in ∆ ADE, \tan θ = \frac{DE}{AE} \Rightarrow \tan θ = \frac{DE}{(\frac{20}{\tan θ})} [Using (i)] \Rightarrow \tan θ = \frac{DE \times \tan θ}{20} \Rightarrow DE = \frac{20 \times \tan θ}{\tan θ} \Rightarrow DE = 20 m Now, CD = DE + CE = 20 + 20 ∴ CD = 40 m$

Hence, the correct answer is option (b).

Disclaimer: The answer given in the textbook is incorrect. The same has been rectified above.

Answer:

Let AB be the lamp post; CD be the girl and DE be her shadow.

We have,

$CD = 1.5 m, AD = 3 m, DE = 4.5 m Let ∠ E = θ In ∆ CDE, \tan θ = \frac{CD}{DE} \Rightarrow \tan θ = \frac{1.5}{4.5} \Rightarrow \tan θ = \frac{1}{3} . . . . . (i) Now, in ∆ ABE, \tan θ = \frac{AB}{AE} \Rightarrow \frac{1}{3} = \frac{AB}{AD + DE} [Using (i)] \Rightarrow \frac{1}{3} = \frac{AB}{3 + 4.5} \Rightarrow AB = \frac{7.5}{3} ∴ AB = 2.5 m$

Hence, the correct answer is option (c).

Question 13:

Let AB be the lamp post; CD be the girl and DE be her shadow.

We have,

$CD = 1.5 m, AD = 3 m, DE = 4.5 m Let ∠ E = θ In ∆ CDE, \tan θ = \frac{CD}{DE} \Rightarrow \tan θ = \frac{1.5}{4.5} \Rightarrow \tan θ = \frac{1}{3} . . . . . (i) Now, in ∆ ABE, \tan θ = \frac{AB}{AE} \Rightarrow \frac{1}{3} = \frac{AB}{AD + DE} [Using (i)] \Rightarrow \frac{1}{3} = \frac{AB}{3 + 4.5} \Rightarrow AB = \frac{7.5}{3} ∴ AB = 2.5 m$

Hence, the correct answer is option (c).

Answer:

Let CD = h be the height of the tower.

We have,

$AB = 2 x, ∠ DAC = 30 ° and ∠ DBC = 45 ° In ∆ BCD, \tan 45 ° = \frac{CD}{BC} \Rightarrow 1 = \frac{h}{BC} \Rightarrow BC = h Now, in ∆ ACD, \tan 30 ° = \frac{CD}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{2 x + h} \Rightarrow 2 x + h = h \sqrt{3} \Rightarrow h \sqrt{3} - h = 2 x \Rightarrow h (\sqrt{3} - 1) = 2 x \Rightarrow h = \frac{2 x}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \Rightarrow h = \frac{2 x (\sqrt{3} + 1)}{(3 - 1)} \Rightarrow h = \frac{2 x (\sqrt{3} + 1)}{2} ∴ h = x (\sqrt{3} + 1) m$

Hence, the correct answer is option (d).

Question 14:

Let CD = h be the height of the tower.

We have,

$AB = 2 x, ∠ DAC = 30 ° and ∠ DBC = 45 ° In ∆ BCD, \tan 45 ° = \frac{CD}{BC} \Rightarrow 1 = \frac{h}{BC} \Rightarrow BC = h Now, in ∆ ACD, \tan 30 ° = \frac{CD}{AC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AB + BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{2 x + h} \Rightarrow 2 x + h = h \sqrt{3} \Rightarrow h \sqrt{3} - h = 2 x \Rightarrow h (\sqrt{3} - 1) = 2 x \Rightarrow h = \frac{2 x}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \Rightarrow h = \frac{2 x (\sqrt{3} + 1)}{(3 - 1)} \Rightarrow h = \frac{2 x (\sqrt{3} + 1)}{2} ∴ h = x (\sqrt{3} + 1) m$

Hence, the correct answer is option (d).

Answer:

Let AB be the rod and BC be its shadow; and $θ$ be the angle of elevation of the sun.

$We have, AB : BC = 1 : \sqrt{3} Let AB = x Then, BC = x \sqrt{3} In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{x}{x \sqrt{3}} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow \tan θ = \tan 30 ° ∴ θ = 30 °$

Hence, the correct answer is option (a).

Question 15:

Let AB be the rod and BC be its shadow; and $θ$ be the angle of elevation of the sun.

$We have, AB : BC = 1 : \sqrt{3} Let AB = x Then, BC = x \sqrt{3} In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{x}{x \sqrt{3}} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow \tan θ = \tan 30 ° ∴ θ = 30 °$

Hence, the correct answer is option (a).

Answer:

Let AB be the pole and BC be its shadow.

We have,

$BC = 2 \sqrt{3} m and ∠ ACB = 60 ° In ∆ ABC, \tan 60 ° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{AB}{2 \sqrt{3}} ∴ AB = 6 m$

Hence, the correct answer is option (b).

Question 16:

Let AB be the pole and BC be its shadow.

We have,

$BC = 2 \sqrt{3} m and ∠ ACB = 60 ° In ∆ ABC, \tan 60 ° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{AB}{2 \sqrt{3}} ∴ AB = 6 m$

Hence, the correct answer is option (b).

Answer:

Let the sun's altitude be $θ$ .

We have,

$AB = 20 m and BC = 20 \sqrt{3} m In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{20}{20 \sqrt{3}} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow \tan θ = \tan 30 ° ∴ θ = 30 °$

Hence, the correct answer is option (a).

Question 17:

Let the sun's altitude be $θ$ .

We have,

$AB = 20 m and BC = 20 \sqrt{3} m In ∆ ABC, \tan θ = \frac{AB}{BC} \Rightarrow \tan θ = \frac{20}{20 \sqrt{3}} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow \tan θ = \tan 30 ° ∴ θ = 30 °$

Hence, the correct answer is option (a).

Answer:

Let AB and CD be the two towers such that AB = x and CD = y.

We have,

$∠ AEB = 30 °, ∠ CED = 60 ° and BE = DE In ∆ ABE, \tan 30 ° = \frac{AB}{BE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{BE} \Rightarrow BE = x \sqrt{3} Also, in ∆ CDE, \tan 60 ° = \frac{CD}{DE} \Rightarrow \sqrt{3} = \frac{y}{DE} \Rightarrow DE = \frac{y}{\sqrt{3}} As, BE = DE \Rightarrow x \sqrt{3} = \frac{y}{\sqrt{3}} \Rightarrow \frac{x}{y} = \frac{1}{\sqrt{3} \times \sqrt{3}} \Rightarrow \frac{x}{y} = \frac{1}{3} ∴ x : y = 1 : 3$

Hence, the correct answer is option (c).

Question 18:

Let AB and CD be the two towers such that AB = x and CD = y.

We have,

$∠ AEB = 30 °, ∠ CED = 60 ° and BE = DE In ∆ ABE, \tan 30 ° = \frac{AB}{BE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{BE} \Rightarrow BE = x \sqrt{3} Also, in ∆ CDE, \tan 60 ° = \frac{CD}{DE} \Rightarrow \sqrt{3} = \frac{y}{DE} \Rightarrow DE = \frac{y}{\sqrt{3}} As, BE = DE \Rightarrow x \sqrt{3} = \frac{y}{\sqrt{3}} \Rightarrow \frac{x}{y} = \frac{1}{\sqrt{3} \times \sqrt{3}} \Rightarrow \frac{x}{y} = \frac{1}{3} ∴ x : y = 1 : 3$

Hence, the correct answer is option (c).

Answer:

(b) $10 \sqrt{3} m$
Let $A B$ be the tower and $O$ be the point of observation.
Also,
∠ $A O B = 30^{o}$ and $O B = 30$ m
Let:
$A B = h$ m

In âˆ† $A O B$ , we have:
$\frac{A B}{O B} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{30} = \frac{1}{\sqrt{3}}$
⇒ $h = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30 \sqrt{3}}{3} = 10 \sqrt{3}$ m

Hence, the height of the tower is $10 \sqrt{3}$ m.

Question 19:

(b) $10 \sqrt{3} m$
Let $A B$ be the tower and $O$ be the point of observation.
Also,
∠ $A O B = 30^{o}$ and $O B = 30$ m
Let:
$A B = h$ m

In âˆ† $A O B$ , we have:
$\frac{A B}{O B} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{30} = \frac{1}{\sqrt{3}}$
⇒ $h = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30 \sqrt{3}}{3} = 10 \sqrt{3}$ m

Hence, the height of the tower is $10 \sqrt{3}$ m.

Answer:

(a) $50 \sqrt{3} m$
Let $A B$ be the string of the kite and $A X$ be the horizontal line.
If $B C$ ⊥ $A X$ , then $A B = 100$ m and ∠ $B A C = 60^{o}$ .
Let:
$B C = h$ m

In the right âˆ† $A C B$ , we have:
$\frac{B C}{A B} = \sin 60^{o} = \frac{\sqrt{3}}{2}$

⇒ $\frac{h}{100} = \frac{\sqrt{3}}{2}$
⇒ $h = \frac{100 \sqrt{3}}{2} = 50 \sqrt{3}$ m
Hence, the height of the kite is $50 \sqrt{3}$ m.

Question 20:

(a) $50 \sqrt{3} m$
Let $A B$ be the string of the kite and $A X$ be the horizontal line.
If $B C$ ⊥ $A X$ , then $A B = 100$ m and ∠ $B A C = 60^{o}$ .
Let:
$B C = h$ m

In the right âˆ† $A C B$ , we have:
$\frac{B C}{A B} = \sin 60^{o} = \frac{\sqrt{3}}{2}$

⇒ $\frac{h}{100} = \frac{\sqrt{3}}{2}$
⇒ $h = \frac{100 \sqrt{3}}{2} = 50 \sqrt{3}$ m
Hence, the height of the kite is $50 \sqrt{3}$ m.

Answer:

(b) $\sqrt{a b}$
Let $A B$ be the tower and $C$ and $D$ be the points of observation on $A C$ .
Let:
∠ $A C B = θ$ , ∠ $A D B = 90 - θ$ and $A B = h$ m
Thus, we have:
$A C = a, A D = b$ and $C D = a - b$

Now, in the right âˆ†ABC, we have:
$\tan θ = \frac{A B}{A C}$ ⇒ $\frac{h}{a} = \tan θ$ ...(i)
In the right âˆ†ABD, we have:
$\tan (90 - θ) = \frac{A B}{A D}$ ⇒ $c o t θ = \frac{h}{b}$ ...(ii)

On multiplying (i) and (ii), we have:
$\tan θ \times c o t θ = \frac{h}{a} \times \frac{h}{b}$
⇒ $\frac{h}{a} \times \frac{h}{b} = 1$ [ âˆµ $\tan θ = \frac{1}{c o t θ}$ ]
⇒ $h^{2} = a b$
⇒ $h = \sqrt{a b}$ m

Hence, the height of the tower is $\sqrt{a b}$ m.

Question 21:

(b) $\sqrt{a b}$
Let $A B$ be the tower and $C$ and $D$ be the points of observation on $A C$ .
Let:
∠ $A C B = θ$ , ∠ $A D B = 90 - θ$ and $A B = h$ m
Thus, we have:
$A C = a, A D = b$ and $C D = a - b$

Now, in the right âˆ†ABC, we have:
$\tan θ = \frac{A B}{A C}$ ⇒ $\frac{h}{a} = \tan θ$ ...(i)
In the right âˆ†ABD, we have:
$\tan (90 - θ) = \frac{A B}{A D}$ ⇒ $c o t θ = \frac{h}{b}$ ...(ii)

On multiplying (i) and (ii), we have:
$\tan θ \times c o t θ = \frac{h}{a} \times \frac{h}{b}$
⇒ $\frac{h}{a} \times \frac{h}{b} = 1$ [ âˆµ $\tan θ = \frac{1}{c o t θ}$ ]
⇒ $h^{2} = a b$
⇒ $h = \sqrt{a b}$ m

Hence, the height of the tower is $\sqrt{a b}$ m.

Answer:

(b) $10 \sqrt{3} m$
Let $A B$ be the tower and $C$ and $D$ be the points of observation such that ∠ $B C D = 30^{o}$ , ∠ $B D A = 60^{o}$ , $C D = 20$ m and $A D = x$ m.

Now, in âˆ† $A D B$ , we have:
$\frac{A B}{A D} = \tan 60^{o}$ $= \sqrt{3}$

⇒ $\frac{A B}{x} = \sqrt{3}$
⇒ $A B = \sqrt{3} x$

In âˆ† $A C B$ , we have:
$\frac{A B}{A C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
$\frac{A B}{20 + x} = \frac{1}{\sqrt{3}}$ ⇒ $A B = \frac{20 + x}{\sqrt{3}}$
∴ $\sqrt{3} x = \frac{20 + x}{\sqrt{3}}$
⇒ $3 x = 20 + x$
⇒ $2 x = 20$ ⇒ $x = 10$
∴ Height of the tower AB = $\sqrt{3} x = 10 \sqrt{3}$ m

Question 22:

(b) $10 \sqrt{3} m$
Let $A B$ be the tower and $C$ and $D$ be the points of observation such that ∠ $B C D = 30^{o}$ , ∠ $B D A = 60^{o}$ , $C D = 20$ m and $A D = x$ m.

Now, in âˆ† $A D B$ , we have:
$\frac{A B}{A D} = \tan 60^{o}$ $= \sqrt{3}$

⇒ $\frac{A B}{x} = \sqrt{3}$
⇒ $A B = \sqrt{3} x$

In âˆ† $A C B$ , we have:
$\frac{A B}{A C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$
$\frac{A B}{20 + x} = \frac{1}{\sqrt{3}}$ ⇒ $A B = \frac{20 + x}{\sqrt{3}}$
∴ $\sqrt{3} x = \frac{20 + x}{\sqrt{3}}$
⇒ $3 x = 20 + x$
⇒ $2 x = 20$ ⇒ $x = 10$
∴ Height of the tower AB = $\sqrt{3} x = 10 \sqrt{3}$ m

Answer:

(c) $16 \sqrt{3} {cm}^{2}$
Let $A B C D$ be the rectangle in which ∠ $B A C = 30^{o}$ and $A C = 8$ cm.

In âˆ† $B A C$ , we have:

$\frac{A B}{A C} = \cos 30^{o} = \frac{\sqrt{3}}{2}$

⇒ $\frac{A B}{8} = \frac{\sqrt{3}}{2}$
⇒ $A B = 8 \frac{\sqrt{3}}{2} = 4 \sqrt{3}$ m
Again,

$\frac{B C}{A C} = \sin 30^{o} = \frac{1}{2}$

⇒ $\frac{B C}{8} = \frac{1}{2}$
⇒ $B C = \frac{8}{2} = 4$ m
∴ Area of the rectangle = $(A B \times B C) = (4 \sqrt{3} \times 4) = 16 \sqrt{3}$ cm²

Question 23:

(c) $16 \sqrt{3} {cm}^{2}$
Let $A B C D$ be the rectangle in which ∠ $B A C = 30^{o}$ and $A C = 8$ cm.

In âˆ† $B A C$ , we have:

$\frac{A B}{A C} = \cos 30^{o} = \frac{\sqrt{3}}{2}$

⇒ $\frac{A B}{8} = \frac{\sqrt{3}}{2}$
⇒ $A B = 8 \frac{\sqrt{3}}{2} = 4 \sqrt{3}$ m
Again,

$\frac{B C}{A C} = \sin 30^{o} = \frac{1}{2}$

⇒ $\frac{B C}{8} = \frac{1}{2}$
⇒ $B C = \frac{8}{2} = 4$ m
∴ Area of the rectangle = $(A B \times B C) = (4 \sqrt{3} \times 4) = 16 \sqrt{3}$ cm²

Answer:

(b) $\frac{1}{2} (\sqrt{3} + 1) km$
Let $A B$ be the hill making angles of depression at points $C$ and $D$ such that ∠ $A D B = 45^{o}$ , ∠ $A C B = 30^{o}$ and $C D = 1$ km.
Let:
$A B = h$ km and $A D = x$ km

In âˆ† $A D B$ , we have:
$\frac{A B}{A D} = \tan 45^{o} = 1$

⇒ $\frac{h}{x} = 1$ ⇒ $h = x$ ...(i)
In âˆ† $A C B$ , we have:
$\frac{A B}{A C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x + 1} = \frac{1}{\sqrt{3}}$ ...(ii)
On putting the value of $h$ taken from (i) in (ii), we get:
$\frac{h}{h + 1} = \frac{1}{\sqrt{3}}$
⇒ $\sqrt{3} h = h + 1$
⇒ $(\sqrt{3} - 1) h = 1$
⇒ $h = \frac{1}{(\sqrt{3} - 1)}$
On multiplying the numerator and denominator of the above equation by $(\sqrt{3} + 1)$ , we get:
$h = \frac{1}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)}{3 - 1} = \frac{(\sqrt{3} + 1)}{2}$ $= \frac{1}{2} (\sqrt{3} + 1)$ km

Hence, the height of the hill is $\frac{1}{2} (\sqrt{3} + 1)$ km.

Question 24:

(b) $\frac{1}{2} (\sqrt{3} + 1) km$
Let $A B$ be the hill making angles of depression at points $C$ and $D$ such that ∠ $A D B = 45^{o}$ , ∠ $A C B = 30^{o}$ and $C D = 1$ km.
Let:
$A B = h$ km and $A D = x$ km

In âˆ† $A D B$ , we have:
$\frac{A B}{A D} = \tan 45^{o} = 1$

⇒ $\frac{h}{x} = 1$ ⇒ $h = x$ ...(i)
In âˆ† $A C B$ , we have:
$\frac{A B}{A C} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x + 1} = \frac{1}{\sqrt{3}}$ ...(ii)
On putting the value of $h$ taken from (i) in (ii), we get:
$\frac{h}{h + 1} = \frac{1}{\sqrt{3}}$
⇒ $\sqrt{3} h = h + 1$
⇒ $(\sqrt{3} - 1) h = 1$
⇒ $h = \frac{1}{(\sqrt{3} - 1)}$
On multiplying the numerator and denominator of the above equation by $(\sqrt{3} + 1)$ , we get:
$h = \frac{1}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)}{3 - 1} = \frac{(\sqrt{3} + 1)}{2}$ $= \frac{1}{2} (\sqrt{3} + 1)$ km

Hence, the height of the hill is $\frac{1}{2} (\sqrt{3} + 1)$ km.

Answer:

(c) $10 \sqrt{3} m$
Let $A B$ be the pole and $A C$ and $A D$ be its shadows.
We have:
∠ $A C B = 30^{o}$ , ∠ $A D B = 60^{o}$ and $A B = 15$ m

In âˆ† $A C B$ , we have:
$\frac{A C}{A B} = c o t 30^{o} = \sqrt{3}$
⇒ $\frac{A C}{15} = \sqrt{3}$ ⇒ $A C = 15 \sqrt{3}$ m

Now, in âˆ† $A D B$ , we have:
$\frac{A D}{A B} = c o t 60^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{A D}{15} = \frac{1}{\sqrt{3}}$ ⇒ $A D = \frac{15}{\sqrt{3}}$ $= \frac{15 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{15 \sqrt{3}}{3} = 5 \sqrt{3}$ m

∴ Difference between the lengths of the shadows = $A C - A D = 15 \sqrt{3} - 5 \sqrt{3} = 10 \sqrt{3}$ m

Question 25:

(c) $10 \sqrt{3} m$
Let $A B$ be the pole and $A C$ and $A D$ be its shadows.
We have:
∠ $A C B = 30^{o}$ , ∠ $A D B = 60^{o}$ and $A B = 15$ m

In âˆ† $A C B$ , we have:
$\frac{A C}{A B} = c o t 30^{o} = \sqrt{3}$
⇒ $\frac{A C}{15} = \sqrt{3}$ ⇒ $A C = 15 \sqrt{3}$ m

Now, in âˆ† $A D B$ , we have:
$\frac{A D}{A B} = c o t 60^{o} = \frac{1}{\sqrt{3}}$
⇒ $\frac{A D}{15} = \frac{1}{\sqrt{3}}$ ⇒ $A D = \frac{15}{\sqrt{3}}$ $= \frac{15 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{15 \sqrt{3}}{3} = 5 \sqrt{3}$ m

∴ Difference between the lengths of the shadows = $A C - A D = 15 \sqrt{3} - 5 \sqrt{3} = 10 \sqrt{3}$ m

Answer:

(b) 30 m
Let $A B$ be the observer and $C D$ be the tower.

Draw $B E$ ⊥ $C D$ , Let $C D = h$ metres. Then,
$A B = 1.5$ m , $B E = A C = 28.5$ m and ∠ $E B D = 45^{o}$ .
$D E = (C D - E C)$ = $(C D - A B) = (h - 1.5)$ m.
In right âˆ† $B E D$ , we have:
$\frac{D E}{B E} = \tan 45^{o} = 1$

⇒ $\frac{(h - 1.5)}{28.5} = 1$

⇒ $h - 1.5 = 28.5$
⇒ $h = 28.5 + 1.5 = 30$ m
Hence the height of the tower is $30$ m.

RS Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 14 - Heights And Distances

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