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Page No 177:

Question 1:

The length of shadow of a tower on the plane ground is 3 times the height of the tower. The angle of elevation of sun is

(a) 45°                       (b) 30°                       (c) 60°                       (d) 90°                               [CBSE 2012]

Answer:



Let the angle of elevation of the sun be θ.

Suppose AB is the height of the tower and BC is the length of its shadow.

It is given that, BC = 3 AB

In right ∆ABC,

tanθ=ABBCtanθ=AB3AB=13tanθ=tan30°θ=30°

Thus, the angle of elevation of the sun is 30º.

Hence, the correct answer is option B.

Page No 177:

Question 2:

The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is 30°. The distance of the car from the base of the tower (in metres) is

(a) 253                    (b) 503                    (c) 753                    (d) 150                               [CBSE 2013]

Answer:



Suppose AB is the tower and C is the position of the car from the base of the tower.

It is given that, AB = 75 m

Now, ACB = CAD = 30°         (Alternate angles)

In right ∆ABC,

tan30°=ABBC13=75 mBCBC=753 m

Thus, the distance of the car from the base of the tower is 753 m.

Hence, the correct answer is option C.

Page No 177:

Question 3:

A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is

(a) 153m                            (b) 1532m                           (c) 152 m                           (d) 15 m                       [CBSE 2013]

Answer:



Suppose AB is the wall and AC is the ladder.

It is given that, AC = 15 m and CAB = 60°.

In right ∆ABC,

cos60°=ABAC12=AB15AB=152 m

Thus, the height of the wall is 152 m.

Hence, the correct answer is option C.

Page No 177:

Question 4:

The angle of depression of a car parked on the road from the top of a 150 m high tower is 30º. The distance of the car from the tower (in metres) is

(a) 503                           (b) 1503                          (c) 1502                          (d) 75                                  [CBSE 2014]

Answer:



Suppose AB is the tower and C is the position of the car from the base of the tower.

It is given that, AB = 150 m

Now, ACB = CAD = 30°         (Alternate angles)

In right ∆ABC,

tan30°=ABBC13=150BCBC=1503 m

Thus, the distance of the car from the tower is 1503 m.

Hence, the correct answer is option B.

Page No 177:

Question 5:

The height of the vertical pole is 3 times the length of its shadow on the ground, then angle of elevation of the sun at that time is

(a) 30º                              (b) 60º                              (b) 45º                              (b) 75º                              [CBSE 2014]

Answer:



Let the angle of elevation of the sun be θ.

Suppose AB is the height of the pole and BC is the length of its shadow.

It is given that, AB = 3 BC

In right ∆ABC,

tanθ=ABBCtanθ=3BCBC=3tanθ=tan60°θ=60°

Thus, the angle of elevation of the sun is 60º.

Hence, the correct answer is option B.



Page No 178:

Question 6:

The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45º. Then the height of the tower (in metres) is

(a) 503                          (b)  50                        (c) 502                        (d) 503                     [CBSE 2014]

Answer:



Suppose AB is the tower and C is a point on the ground.

It is given that, BC = 50 m and ACB = 45°.

In right ∆ABC,

tan45°=ABBC1=AB50AB=50 m

Thus, the height of the tower is 50 m.

Hence, the correct answer is option B.

Page No 178:

Question 7:

A ladder makes an angle of 60º with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder (in metres) is

(a) 43                          (b) 43                          (c) 22                          (d) 4                          [CBSE 2014]

Answer:



Suppose AC is the ladder and BC is the distance of the foot of the ladder from the wall.

It is given that, BC = 2 m and ACB = 60°.

In right ∆ABC,

cos60°=BCAC12=2 ACAC=4 m

Thus, the length of the ladder is 4 m.

Hence, the correct answer is option D.

Page No 178:

Question 8:

Mark the correct alternative in each of the following:

The ratio of the length of a rod and its shadow is 1 : 3. The angle of elevation of the sum is

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Answer:

Let be angle of elevation of sun.

Given that: length of road and its shadow

Here, we have to find angle of elevation of sun.

So we use trigonometric ratios.

In a triangle,

Hence the correct option is .

Page No 178:

Question 9:

If the angle of elevation of a tower from a distance of 100 metres from its foot is 60°, then the height of the tower is

(a) 1003 m

(b) 1003 m

(c) 50 3

(d) 2003 m

Answer:

Let be the height of tower is meters

Given that: angle of elevation is from tower of foot and distance meters.

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

Hence correct option is .

Page No 178:

Question 10:

If the altitude of the sum is at 60°, then the height of the vertical tower that will cast a shadow of length 30 m is

(a) 103 m

(b) 15 m

(c) 303 m

(d) 152 m

Answer:

Let be the height of vertical tower

Given that: altitude of sun is and shadow of length meters.

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

Hence the correct option is .

Page No 178:

Question 11:

If the angles of elevation of a tower from two points distant a and b (a>b) from its foot and in the same straight line from it are 30° and 60°, then the height  of the tower is

(a) a+b

(b) ab

(c) a-b

(d) ab

Answer:

Let be the height of tower

Given that: angle of elevation are and.

Distance and

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,

Put

Hence the correct option is .

Page No 178:

Question 12:

If the angles of elevation of the top of a tower from two points distant a and b from the  base and in the same straight line with it are complementary, then the height of the tower is

(a) ab

(b) ab

(c) ab

(d) ab

Answer:

Let be the height of tower.

Given that: angle of elevation of top of the tower are and .

Distance and

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

 

Again in a triangle ABD,

Put

Hence the correct option is .

Page No 178:

Question 13:

From a light house the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the light house is h metres, the distance between the ships is

(a) 3+ 1 h metres

(b) 3- 1 h metres

(c) 3 h metres

(d) 1 + 1+13 h metres

Answer:

Let the height of the light house AB be meters

Given that: angle of depression of ship are and.

Distance of the ship C = and distance of the ship D =

Here, we have to find distance between the ships.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,

Now, distance between the ships

Hence the correct option is .

Page No 178:

Question 14:

The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance d towards the foot of the tower the angle of elevation is found to be β. The height of the tower is

(a) dcot α+cot β

(b) dcot α-cot β

(c) dtan β-tan α 

(d) dtan β+tan α 

Answer:

The given information can be represented with the help of a diagram as below.

Here, CD = h is the height of the tower. Length of BC is taken as x.

 

In ΔACD,

In ΔBCD.

 

From (1) and (2),

Hence the correct option is .

Page No 178:

Question 15:

The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, then the length of the wire is

(a) 12 m

(b) 10 m

(c) 8 m

(d) 6 m

Answer:

Let be the length of wire.

Given that wire makes an angle

Now, ,

Here, we have to find length of wire.

So we use trigonometric ratios.

In a triangle,

Hence the correct option is .

Page No 178:

Question 16:

From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is

(a) 25 m

(b) 50 m

(c) 75 m

(d) 100 m

Answer:

Given that: height of cliff is m and angle of elevation of the tower is equal to angle of depression of foot of the tower that is.

Now, the given situation can be represented as,

Here, D is the top of cliff and BE is the tower.

Let CE = h, . Then, = = x

 

Here, we have to find the height of the tower BE.

So, we use trigonometric ratios.

In a triangle ABD,

Again in a triangle,

Thus, height of the tower = BE = BC + CE = (25 + 25) m = 50 m

Hence, the correct option is .

Page No 178:

Question 17:

The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart. the height of the light house is

(a) 503+1 m

(b) 503-1 m

(c) 50 3-1 m

(d) 50 3+1 m

Answer:

Let = be the light house.

The given situation can be represented as,

It is clear that and

Again, let and m is given.

Here, we have to find the height of light house.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,,

Put

Hence the correct option is .

Page No 178:

Question 18:

If the angle of elevation of a cloud from a point 200 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°, then the height of the cloud above the lake is

(a) 200 m

(b) 500 m

(c) 30 m

(d) 400 m

Answer:

Let be the surface of the lake and be the point of observation. So m.

The given situation can be represented as,

Here, is the position of the cloud and is the reflection in the lake. Then .

Let be the perpendicular from on . Then and .

Let , , then and

Here, we have to find the height of cloud.

So we use trigonometric ratios.

In ,

Again in,

Put

Now,

Hence the correct answer is .



Page No 179:

Question 19:

The height of a tower is 100 m. When the angle of elevation of the sun changes from 30° to 45°, the shadow of the tower becomes x metres less. The value of x is

(a) 100 m

(b) 1003 m

(c) 1003 -1 m

(d) 1003m

Answer:

The given situation can be represented as,

Here, AB is the tower of height meters.

When angle of elevation of sun changes from to , .

We assumed that

Here we have to find the value of

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,

Put

Hence the correct option is .

Page No 179:

Question 20:

Two persons are a metres apart and the height of one is double that of the other. If from the middle point of the line joining their feet, an observer finds the angular elevation of their tops to be complementary, then the height of the shorter post is

(a) a4

(b) a2

(c) a2

(d) a22

Answer:

Let AB and CD be the two persons such that AB < CD.

Then, let AB = h so that CD = 2h

Now, the given information can be represented as,

Here, E is the midpoint of BD.

We have to find height of the shorter person.

So we use trigonometric ratios.

In triangle ECD,

Again in triangle ABE,

Hence the correct option is .

Page No 179:

Question 21:

The angle of elevation of a cloud from a point h metre above a lake is θ. The angle of depression of its reflection in the lake is 45°. The height of the cloud is

(a) h tan (45° + θ)

(b) h cot (45° − θ)

(c) h tan (45° − θ)

(d) h cot (45° + θ)

Answer:

Let be the surface of the lake and be the point of observation. So .

The given situation can be represented as,

Here, is the position of the cloud and is the reflection in the lake. Then .

Let be the perpendicular from on . Then and .

Let , , then and

Here, we have to find the height of cloud.

So we use trigonometric ratios.

In ,

Again in,

Now,

Hence the correct answer is .

Page No 179:

Question 22:

A tower subtends an angle of 30° at a point on the same level as its foot. At a second point h metres above the first, the depression of the foot of the tower is 60°. The height of the tower is

(a) h2 m

(b) 3h m

(c) h3 m

(d) h3m

Answer:

Let AB be the tower and C is a point on the same level as its foot such that ACB = 30°

The given situation can be represented as,

Here D is a point h m above the point C.

In ΔBCD,

Again in triangle ABC,

Hence the correct option is .

Page No 179:

Question 23:

It is found that on walking x meters towards a chimney in a horizontal line through its base, the elevation of its top changes from 30° to 60°. The height of the chimney is

(a) 32x

(b) 23x

(c) 32x

(d) 23x

Answer:

Let be the height of chimney

Given that: angle of elevation changes from angle to .

Then Distance becomes and we assume

Here, we have to find the height of chimeny.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,

Hence the correct option is .

Page No 179:

Question 24:

The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30°than when it was 45°. The height of the tower in metres is

(a) 3+1 x

(b) 3-1 x

(c) 23x

(d) 32x

Answer:

 

Let be the height of tower

Given that: angle of elevation of sun are and.

Then Distance and we assume

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

Again in a triangle ABD,

Hence the correct option is .

Page No 179:

Question 25:

Two poles are 'a' metres apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the smaller is

(a) 2a metres

(b) a22metres

(c) a2 metres

(d) 2a metres

Answer:

Let AB and CD be the two posts such that AB < CD.

Then, let AB = h so that CD = 2h

Now, the given information can be represented as,

Here, E is the midpoint of BD.

We have to find height of the shorter post.

So we use trigonometric ratios.

In triangle ECD,

Again in triangle ABE,

Hence the correct option is .

Page No 179:

Question 26:

The tops of two poles of height 16 m and 10 m are connected by a wire of length l metres. If the wire makes an angle of 30° with the horizontal, then l =

(a) 26

(b) 16

(c) 12

(d) 10

Answer:

Let AB and CD be the poles such that AB = 16 m and CD = 10 m.

The given information can be represented as

Here, AC is the length of wire which is .

Also, AE = ABBE = 16 m − 10 m = 6 m

We have to find the length of wire .

So we use trigonometric ratios.

In triangle ACE,

Hence the correct option is .

Page No 179:

Question 27:

If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is

(a) 1.5 m

(b) 2 m

(c) 2.5 m

(d) 2.8 m

Answer:

Let AB be the lamp post and CD =1.5 m be the girl.

The given information can be represented as

Here, shadow of girl is DE = 4.5 m and BD = 3 m.

In ΔCDE,

In ΔABE,

Therefore, height of the lamp post is 2.5 m

Hence the correct option is .

Page No 179:

Question 28:

We know that during vacation period many people love to go out of the city and gain some experience about historical and scientific values. Keeping in views, Mr. Ramlal decided to go somewhere out of the country and chosen the country USA. In the series of sight seeing, he has first chosen the place sky tower of Mexico City. Then he decided to stand on a building and wanted to see the sky tower. Mr. Ramlal whose height is 2.3 m stood on the top of a building and started to look at the top of sky tower. The horizontal distance between sky tower and the building is 120 mt. as shown in the given figure. The angle of elevation of the top and angle of depression of the bottom of the sky tower are 60° and 30° respectively. Looking into the above circumstances to give the answer the following questions:


(i) What is the height of the building excluding the height of Mr. Ramlal standing on it?

(a) 65.98 m
(b) 66.98 m
(c) 67.98 m
(d) 68.98 m

(ii) Find the height of the building including height of Mr. Ramlal?
(a) 303 m
(b) 353 m
(c) 403 m
(d) 453 m​

(iii) What is the length of line of sight of Mr. Ramlal to the base of the Sky tower?
(a) 403 m
(b) 803 m
(c) 1003 m
(d) 1203 m​


(iv) Find the distance from the eye of Mr. Ramlal to top of the sky tower along the line of sight?
(a) 210 m
(b) 220 m
(c) 230 m
(d) 240 m

(v) Find the height of the sky tower?
(a) 1203 m
(b) 1403 m
(c) 1503 m
(d) 1603 m​​

Answer:

Let AB be the sky tower, CD be the height of Mr. Rampal, DE be the height of the building.



(i) In ∆CEB,
tan30°=CEEB13=CE120CE=1203CE=403 m
DE=CE-CD=403-2.3=69.28-2.3=66.98 m

Hence, the correct answer is option (b).

(ii) In ∆CEB,
tan30°=CEEB13=CE120CE=1203CE=403 m

Hence, the correct answer is option (c).

(iii) In ∆CEB,
cos30°=EBCB32=120CBCB=2403CB=803 m


Hence, the correct answer is option (b).

(iv) In ∆AFC,
cos60°=CFAC12=120ACAC=240 m

Hence, the correct answer is option (d).

(v) In ∆AFC,
tan60°=AFCF3=AF120AF=1203 m
AB=AF+FBAB=1203+403  FB=CEAB=1603 m

Hence, the correct answer is option (d).



Page No 180:

Question 29:

A helicopter lifts up 1000 feet over an island and spots a swimmer that need to be rescued. Using a distant land mark, the helicopter pilot determines the angle of depression.


(i) As the angle of depression increases what will be the effect?

(a) The helicopter gets further from the island.
(b) The helicopter gets closer to the island.
(c) The swimmer gets closer to the island.
(d) The swimmer gets further from the island.

(ii) How does the swimmer's distance from island changes as the angle of depression is halved from 60° to 30°?
(a) The swimmer's distance decreases to less than a quarter of his starting distance.
(b) The swimmer's distance from the island doubles.
(c) The swimmer's distance from the island increases three times.
(d) The swimmer's distance from the island is halved.

(iii) For which angle of depression both the helicopter and swimmer's will be at same distance?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

(iv) Let the swimmer start out 1019 ft. from the island. If he swims half of the distance, what is angle of depression?
(a) nearly 30°
(b) nearly 45°
(c) nearly 60°
(d) nearly 90°

(v) How would the angle of depression be affected if the helicopter left its initial position and moved vertically upward?
(a) angle of depression doesn't change
(b) angle of depression increases
(c) angle of depression decreases
(d) none of the above

Answer:


(i) If the angle of the depression increases then, the value of the θ will decrease.
As the value of the θ will decrease, and then the helicopter gets farther from the island.

Hence, the correct answer is option (c).

(ii)
tan60°=1000X3=1000XX=10003

When the angle of depression is halved from 60° to 30°

tan30°=1000X13=1000XX=10003X=3×10003
Thus, the swimmer's distance from the island increases three times.

Hence, the correct answer is option (c).

(iii) Let the angle of depression when both the helicopter and swimmer's will be at same distance be Ï•.
Then X = 1000 ft        (∵ In a triangle sides corresponding to the equal angles are also equal)

tanϕ=10001000tanϕ=1tanϕ=tan45°ϕ=45°

Hence, the correct answer is option (b).

(iv) The distance between the Island and the swimmer, when he swims half of the 1019 ft, i.e. 510 ft.

tanθ=1000510tanθ=1.96tanθtan60°θ60°

Hence, the correct answer is option (c).

(v) If the helicopter left its initial position and moved vertically upward, then the value of the θ will decrease.
As the value of the θ will decreases, then the angle of depression increases.

Hence, the correct answer is option (b).



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