Mathematics Part II Solutions Solutions for Class 10 Math Chapter 6 Trigonometry are provided here with simple step-by-step explanations. These solutions for Trigonometry are extremely popular among Class 10 students for Math Trigonometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 10 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 131:

Question 1:

If sinθ=725 , find the values of cosθ and tan​θ.

Answer:


We have,

sin2θ+cos2θ=17252+cos2θ=1cos2θ=1-49625=625-49625=576625cosθ=576625=2425
Now,

tanθ=sinθcosθtanθ=7252425tanθ=724
Thus, the values of cosθ and tanθ are 2425 and 724, respectively.

Page No 131:

Question 2:

If tanθ=34, find the values of sec​θ and cos​θ

Answer:


We have,

sec2θ=1+tan2θsec2θ=1+342sec2θ=1+916=16+916=2516secθ=2516=54
Now,

cosθ=1secθcosθ=154cosθ=45
Thus, the values of sec​θ and cos​θ are 54 and 45, respectively.

Page No 131:

Question 3:

If cotθ=409 , find the values of cosecθ and sinθ.

Answer:


We have,

cosec2θ=1+cot2θcosec2θ=1+4092cosec2θ=1+160081=168181cosecθ=168181=419
Now,

sinθ=1cosecθsinθ=1419sinθ=941
Thus, the values of cosecθ and sinθ are 419 and 941, respectively.

Page No 131:

Question 4:

If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ and sinθ.

Answer:


5secθ-12cosecθ=05secθ=12cosecθ5cosθ=12sinθsinθcosθ=125tanθ=125
We have,

sec2θ=1+tan2θsec2θ=1+1252sec2θ=1+14425=16925secθ=16925=135
Now,

cosθ=1secθcosθ=1135cosθ=513
Also,

sinθcosθ=tanθsinθ=tanθ×cosθsinθ=125×513=1213
Thus, the values of secθ, cosθ and sinθ are 135513 and 1213, respectively.

Page No 131:

Question 5:

If tanθ = 1 them, find the values of sinθ+cosθsecθ+cosecθ .

Answer:


tanθ = 1

We know that, tan45º = 1

∴ θ = 45º

Now,

sin45°=12cos45°=12sec45°=2cosec45°=2
sinθ+cosθsecθ+cosecθ=sin45°+cos45°sec45°+cosec45°=12+122+2=2222=12

Page No 131:

Question 6:

Prave that:

(1) sin2θcosθ+cosθ=secθ

(2) cos2θ1+tan2θ=1

(3) 1-sinθ1+sinθ=secθ-tanθ

(4) secθ-cosθcotθ+tanθ=tanθ secθ

(5) cotθ+tanθ=cosecθ secθ

(6) 1secθ-tanθ=secθ+tanθ

(7) sec4θ-cos4θ=1-2cos2θ

(8) secθ+tanθ=cosθ1-sinθ

(9) If tanθ+1tanθ=2, then show that tan2θ+1tan2θ=2

(10) tanA1+tan2A2+cotA1+cot2A2=sin A cos A

(11) sec4A1-sin4A-2tan2A=1

(12) tanθsecθ-1=tanθ+secθ+1tanθ+secθ-1

Answer:


(1) 
sin2θcosθ+cosθ=sin2θ+cos2θcosθ=1cosθ=secθ
(2)
cos2θ1+tan2θ=cos2θ×sec2θ=cos2θ×1cos2θ=1
(3)
1-sinθ1+sinθ=1-sinθ1+sinθ×1-sinθ1-sinθ=1-sinθ21-sin2θ=1-sinθ2cos2θ                cos2θ+sin2θ=1
=1-sinθcosθ=1cosθ-sinθcosθ=secθ-tanθ
(4)
secθ-cosθcotθ+tanθ=1cosθ-cosθcosθsinθ+sinθcosθ=1-cos2θcosθsin2θ+cos2θsinθcosθ=sin2θcosθ×1sinθcosθ                sin2θ+cos2θ=1=sinθcosθ×1cosθ=tanθsecθ
(5)
cotθ+tanθ=cosθsinθ+sinθcosθ=sin2θ+cos2θsinθcosθ=1sinθcosθ                sin2θ+cos2θ=1=1sinθ×1cosθ=cosecθsecθ
(6)
1secθ-tanθ=1secθ-tanθ×secθ+tanθsecθ+tanθ=secθ+tanθsec2θ-tan2θ                 a+ba-b=a2-b2=secθ+tanθ                      1+tan2θ=sec2θ
(7)
Disclaimer: There is printing mistake in the question. The correct question should be sin4θ-cos4θ=1-2cos2θ. The solution has been provided accordingly.

sin4θ-cos4θ=sin2θ2-cos2θ2=sin2θ-cos2θsin2θ+cos2θ                  a2-b2=a+ba-b=sin2θ-cos2θ                                               sin2θ+cos2θ=1=1-cos2θ-cos2θ=1-2cos2θ
(8)
secθ+tanθ=1cosθ+sinθcosθ=1+sinθcosθ=cosθ1+sinθcos2θ=cosθ1+sinθ1-sin2θ                sin2θ+cos2θ=1
=cosθ1+sinθ1+sinθ1-sinθ=cosθ1-sinθ
(9)
tanθ+1tanθ=2

Squaring on both sides, we get

tanθ+1tanθ2=22tan2θ+1tan2θ+2×tanθ×1tanθ=4tan2θ+1tan2θ+2=4tan2θ+1tan2θ=4-2=2
(10)
tanA1+tan2A2+cotA1+cot2A2=tanAsec2A2+cotAcosec2A2                1+tan2θ=sec2θ and 1+cot2θ=cosec2θ =sinAcosA×cos4A+cosAsinA×sin4A     cosθ=1secθ and sinθ=1cosecθ=sinAcos3A+cosAsin3A
=sinAcosAcos2A+sin2A=sinAcosA                                     cos2θ+sin2θ=1
(11)
sec4A1-sin4A-2tan2A=sec4A-sec4Asin4A-2tan2A=1+tan2A2-sin4Acos4A-2tan2A                  sec2θ=1+tan2θ and secθ=1cosθ=1+tan4A+2tan2A-tan4A-2tan2A         a+b2=a2+b2+2ab=1
(12)
tanθ+secθ+1tanθ+secθ-1=tanθ+secθ+1tanθ+secθ-1×tanθ+secθ+1tanθ+secθ+1=tan2θ+sec2θ+1+2tanθ+2secθ+2secθtanθtan2θ+sec2θ+2tanθsecθ-1=2sec2θ+2tanθ+2secθ+2secθtanθ2tan2θ+2tanθsecθ               1+tan2θ=sec2θ=2tanθ+secθ+2secθtanθ+secθ2tanθtanθ+secθ
=2tanθ+secθsecθ+12tanθtanθ+secθ=secθ+1tanθ=secθ+1tanθ×secθ-1secθ-1=sec2θ-1tanθsecθ-1               a-ba+b=a2-b2=tan2θtanθsecθ-1=tanθsecθ-1



Page No 137:

Question 1:

A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.

Answer:


Let AB be the church and C be the position of the person from the church.  

Suppose the height of the church be h m.



In right ∆ABC,

tan45°=ABBC1=h80h=80 m
Thus, the height of the church is 80 m.

Page No 137:

Question 2:

From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. 3=1.73

Answer:


Let AB be the lighthouse and C be the position of the ship from the lighthouse.

Suppose the distance of the ship from the lighthouse be x m.



In right ∆ABC,

tan60°=ABBC3=90xx=903=303x=30×1.73=51.9 m
Thus, the ship is 51.9 m away from the lighthouse.

Page No 137:

Question 3:

Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building ?

Answer:

Let AB and CD be the two building standing on the road.

Suppose the height of the second building be h m.



Here, CD = 10 m, BD = 12 m and CE ⊥ AB. 

AE = AB − EB = (h − 10) m            (EB = CD)

CE = BD = 12 m

In right ∆AEC,

tan60°=AECE3=h-1012h-10=123h=10+123 m
Thus, the height of the second building is 10+123 m.

Page No 137:

Question 4:

Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops in 22 metre. Find the angle made by the wire with the horizontal.

Answer:

Let AB and CD be the two poles standing on the ground. 

Suppose the angle made by the wire with the horizontal be θ.



Here, AB = 18 m and CD = 7 m.

Length of the wire fastened at their tops = AC = 22 m

AE = AB − EB = 18 − 7 = 11 m            (EB = CD)

In right ∆AEC,

sinθ=AEACsinθ=1122=12sinθ=12=sin30°θ=30°
Thus, the angle made by the wire with the horizontal is 30º.

Page No 137:

Question 5:

A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.

Answer:




Let AC be the original height of the tree. Suppose BD be the broken part of the tree which is rested at D from the base of the tree. 

Here, CD = 20 m and ∠BDC = 60º.

In right ∆BCD,

tan60°=BCCD3=BC20BC=203 m          .....1
Also,

cos60°=CDBD12=20BDBD=40 m          .....2
∴ Height of the tree = AB + BC = BD + BC = 40+203 m               [Using (1) and (2)]

Thus, the height of the tree is 40+203 m.

Page No 137:

Question 6:

A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. 3=1.73

Answer:

Let AB be the height of kite above the ground and C be the position of the string attached to the kite which is tied at the ground.

Suppose the length of the string be x m.



Here, AB = 60 m and ∠ACB = 60º

In right ∆ABC,

sin60°=ABAC32=60xx=1203=403x=40×1.73=69.2 m
Thus, the length of the string is 69.2 m.



Page No 138:

Question 1:

Choose the correct alternative answer for the following questions.


(1) sinθ cosecθ = ?
(A) 1

(B) 0

(C) 12

(D) 2

(2) cosec 45° =?
(A) 12

(B) 2

(C)  32

(D)23
 
(3) 1 + tan2 θ  = ?
(A) cot2 θ
(B) cosec2 θ
(C) sec2 θ
(D) tan2 θ 

(4) When we see at a higher level , from the horizontal line,angle formed is ....... .
(A) angle of elevation.
(B) angle of depression.
(C) 0
(D) straight angle.

Answer:

(1)
sinθcosecθ=sinθ×1sinθ=1
Hence, the correct answer is option (A).

(2)
cosec45°=2

Hence, the correct answer is option (B).

(3)
1+tan2θ=sec2θ

Hence, the correct answer is option (C).

(4)
When we see at a higher level, from the horizontal line, angle formed is angle of elevation.



Hence, the correct answer is option (A).

Page No 138:

Question 2:

If sinθ=1161, find the values of cosθ using trigonometric identity.

Answer:


We have,

sin2θ+cos2θ=1cos2θ=1-sin2θcos2θ=1-11612cos2θ=1-1213721=3721-1213721=36003721
cosθ=60612=6061

Thus, the value of cosθ is 6061.

Page No 138:

Question 3:

If tanθ = 2, find the values of other trigonometric ratios.

Answer:


tanθ = 2         (Given)

We have,

sec2θ=1+tan2θsecθ=1+tan2θsecθ=1+22secθ=1+4=5
cosθ=1secθ=15

Now,

tanθ=sinθcosθsinθ=tanθ×cosθsinθ=2×15=25cosecθ=1sinθ=125=52
Also,

cotθ=1tanθ=12

Page No 138:

Question 4:

If secθ=1312 , find the values of other trigonometric ratios.

Answer:


secθ=1312              (Given)

cosθ=1secθ=11312=1213
We have,

1+tan2θ=sec2θtanθ=sec2θ-1tanθ=13122-1tanθ=169144-1
tanθ=169-144144=25144tanθ=512cotθ=1tanθ=1512=125
Now,

tanθ=sinθcosθsinθ=tanθ×cosθsinθ=512×1213=513cosecθ=1sinθ=1513=135

Page No 138:

Question 5:

Prove the following.

(1) secθ (1 – sinθ) (secθ + tanθ) = 1

(2) (secθ + tanθ) (1 – sinθ) = cosθ

(3) sec2θ + cosec2θ = sec2θ × cosec2θ

(4) cot2θ – tan2θ = cosec2θ – sec2θ

(5) tan4θ + tan2θ = sec4θ – sec2θ

(6) 11-sinθ+11+sinθ=2 sec2θ

(7) sec6x – tan6x = 1 + 3sec2x × tan2x

(8) tanθsecθ+1=secθ-1tanθ

(9) tan3θ-1tanθ-1=sec2θ+tanθ

(10) sinθ-cosθ+1sinθ+cosθ-1=1sinθ-tanθ

Answer:


(1)
secθ1-sinθsecθ+tanθ=secθ-secθsinθsecθ+tanθ=secθ-sinθcosθsecθ+tanθ=secθ-tanθsecθ+tanθ=sec2θ-tan2θ=1                                        1+tan2θ=sec2θ
(2)
secθ+tanθ1-sinθ=1cosθ+sinθcosθ1-sinθ=1+sinθcosθ1-sinθ=1-sin2θcosθ=cos2θcosθ              sin2θ+cos2θ=1=cosθ
(3)
sec2θ+cosec2θ=1cos2θ+1sin2θ=sin2θ+cos2θcos2θsin2θ=1cos2θsin2θ=1cos2θ×1sin2θ=sec2θcosec2θ
(4)
cot2θ-tan2θ=cosec2θ-1-sec2θ-1                        1+tan2θ=sec2θ & 1+cot2θ=cosec2θ=cosec2θ-1-sec2θ+1=cosec2θ-sec2θ
(5)
tan4θ+tan2θ=sec2θ-12+sec2θ-1                                    1+tan2θ=sec2θ=sec4θ-2sec2θ+1+sec2θ-1                            a-b2=a2-2ab+b2=sec4θ-sec2θ
(6)
11-sinθ+11+sinθ=1+sinθ+1-sinθ1-sinθ1+sinθ=21-sin2θ                                          a-ba+b=a2-b2=2cos2θ                                               sin2θ+cos2θ=1=2sec2θ
(7)
We have,

sec2x-tan2x=1

Cubing on both sides, we get

sec2x-tan2x3=13sec2x3-tan2x3-3×sec2x×tan2x×sec2x-tan2x=1                     a-b3=a3-b3-3aba-bsec6x-tan6x-3sec2xtan2x=1sec6x-tan6x=1+3sec2xtan2x
(8)
tanθsecθ+1=tanθsecθ+1×secθ-1secθ-1=tanθsecθ-1sec2θ-1=tanθsecθ-1tan2θ                        1+tan2θ=sec2θ=secθ-1tanθ
(9)
tan3θ-1tanθ-1=tanθ-1tan2θ+tanθ×1+1tanθ-1                    a3-b3=a-ba2+ab+b2=tan2θ+tanθ+1=sec2θ+tanθ                                                     1+tan2θ=sec2θ
(10)
sinθ-cosθ+1sinθ+cosθ-1
=sinθ-cosθ+1cosθsinθ+cosθ-1cosθ             (Dividing numerator and denominator by cosθ)
=sinθcosθ-cosθcosθ+1cosθsinθcosθ+cosθcosθ-1cosθ
=tanθ-1+secθtanθ+1-secθ=tanθ-1+secθtanθ+sec2θ-tan2θ-secθ                                          1+tan2θ=sec2θ=tanθ-1+secθsecθ-tanθsecθ+tanθ-secθ-tanθ                    a2-b2=a-ba+b     =tanθ-1+secθsecθ-tanθsecθ+tanθ-1=1secθ-tanθ



Page No 139:

Question 6:

A boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.

Answer:

Let AB be the building and C be the position of the boy from the building.  

Suppose the height of the building be h m.



Here, BC = 48 m and ∠ACB = 30º.

In right ∆ABC,

tan30°=ABBC13=h48h=483=163 m
Thus, the height of the building is 163 m.

Page No 139:

Question 7:

From the top of the light house, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the light-house is 100 meters, then find how far the ship is from the light-house.

Answer:

Let AB be the lighthouse and C be the position of the ship from the lighthouse.

Suppose the distance of the ship from the lighthouse be x m.



Here, AB = 100 m and ∠ACB = 30º.

In right ∆ABC,

tan30°=ABBC13=100xx=1003 m
Thus, the ship is 1003 m away from the lighthouse.

Page No 139:

Question 8:

Two buildings are in front of each other on a road of width 15 meters. From the top of the first building, having a height of 12 meter, the angle of elevation of the top of the second building is 30°.What is the height of the second building?

Answer:

Let AB and CD be the two building standing on the road.

Suppose the height of the second building be h m.



Here, AB = 12 m, BD = 15 m, ∠CAE = 30º and AE ⊥ CD. 

CE = CD − ED = (h − 12) m            (ED = AB)

AE = BD = 15 m

In right ∆AEC,

tan30°=CEAE13=h-1215h-12=153=53h=12+53 m
Thus, the height of the second building is 12+53 m.

Page No 139:

Question 9:

A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)

Answer:


Let AB be the platform of the fire brigade van and AD be the ladder. 

Suppose the maximum height from the ground upto which the ladder can reach be h m.



Here, AB = 2 m, AD = 20 m, ∠DAE = 70º and AE ⊥ CD.

DE = CD − CE = (− 2) m             (CE = AB)

In right ∆AED,

sin70°=DEAD0.94=h-220h-2=20×0.94=18.8h=18.8+2=20.8 m
Thus, the maximum height from the ground upto which the ladder can reach is 20.8 m.

Page No 139:

Question 10:

While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing. (sin 20° = 0.342)

Answer:


Let the plane was at a height of h m when it started landing. 

Average speed of the plane = 200 km/h = 200×518=5009 m/s

Time taken by plane to reach the ground = 54 s

∴ Distance covered by the plane to reach the ground = Average speed of the plane × Time taken by plane to reach the ground

5009×54 

= 3000 m

Here, AC = 3000 m and ∠ACB = 20º

In right ∆ABC,

sin20°=ABAC0.342=h3000h=0.342×3000=1026 m
Thus, the plane was at the height of 1026 m when it started landing.



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