# Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.5 Pdf

## Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.5 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.5: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.5 – Trigonometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 6 Exercise 6.5 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 6.5

### TN Samacheer Kalvi 10th Maths Solutions Chapter 6 – Trigonometry

 Board TNSCERT Class 10th Maths Class 10 Subject Maths Chapter 6 (Exercise 6.5) Chapter Name Trigonometry

### TNSCERT Class 10th Maths Pdf | all Exercise Solution

Exercise – 6.5

(*)Multiple choice questions:

(1) The value of sin2θ + 1/1+tan2θ is equal to

(A) tan2θ

(B) 1

(C) cot2θ

(D) 0

Solution:

= sin2 θ + 1/1+tan2 θ

= sin2 θ + 1/sec2 θ

= sin2 θ + cos2 θ

= 1 (B)

(2) tanθ cosec2θ – tanθ is equal to

(A) secθ

(B) cot2θ

(C) sinθ

(D) cotθ

Solution:

= tanθ cosec2θ – tanθ

= tanθ (cosec2θ – 1)

= tanθ. 1/tan2θ

= 1/tanθ = cotθ (D)

(3) If (sinα +cosecα)2 + (cosα + secα)2 = k + tan2α + cot2α, then the value of k is equal to

(A) 9

(B) 7

(C) 5

(D) 3

Solution:

(sinα + cosecα)2 + (cosα + secα)2 = k + tan2α + cot2α

sin2α + 2sinα.cosecα + cosec2α + cos2α + 2cosα.secα + sec2α

= k + tan2α + cot2α

1 + 2 + 2 + cosec2α + sec2α = k + tan2α + cot2α

5 + 1 + cot2α + 1 + tan2α = k + tan2α + cot2α

7 + tan2α + cot2α = k + tan2α + cot2α

∴ k = 7 (B)

(4) If sinθ + cosθ = a and secθ + cosecθ = b, then the value of b (a2 – 1) is equal to

(A) 2a

(B) 3a

(C) 0

(D) 2ab

Solution:

b (a2 – 1)

= (secθ + cosecθ) (sin2θ + cos2θ + 2sinθ cosθ – 1)

= (secθ + cosecθ) (1 + 2sinθ cosθ- 1)

= (secθ + cosecθ) 2 sinθcosθ.

= 1/cosθ × 2sinθ cosθ + 1/sinθ + 2sinθ.cosθ

= 2 sinθ + 2cosθ

= 2 (sinθ + cosθ)

= 2a [∵sinθ + cosθ = a]

= 2a (A)

(5) If 5x = secθ and 5/x tanθ, then x2 – 1/x2 is equal to

(A) 25

(B) 1/25

(C) 5

(D) 1

Solution:

5x = secθ

x = secθ/5

and 5/x = tanθ

1/x = tanθ/

Now, x2 – /x2

= (secθ/5)2 – (tanθ/5)2

= sec2θ – tan2θ/25

= 1/25 (B)

(6) If sinθ = cosθ, then 2tan2θ + sin2θ – 1 is equal to

(A) -3/2

(B) 3/2

(C) 2/3

(D) -2/3

Solution:

Sinθ = cosθ (Given)

2tan2θ + sin2θ – 1

sinθ = cosθ

∴ sin45° = cos45°

∴θ = 45°

= 2 sin2θ/cos2θ + sin2θ – 1

= 2 (sinθ/cosθ)2 + sin2θ – 1

= 2 (sinθ/sinθ)2 + sin2θ – 1 [∵sinθ = cosθ]

= 2 + sin2θ – 1

= 1 + sin2θ

= 1 + (sin45°)2

= 1 + 1/2 = 3/2 (B)

(7) If x = a tanθ and y = bsecθ then

(A) y2/b2 – x2/a2 = 1

(B) x2/a2 – y2/b2 = 1

(C) x2/a2 + y2/b2 = 1

(D) x2/a2 – y2/b2 = 0

Solution:

(A) x = atanθ and y = b secθ

= y2/b2 – x2/a2

= b2 sec2θ/b2 – a2tan2θ/a2

= sec2θ – tan2θ

= 1

(8) (1 + tanθ + secθ) (1 + cotθ – cosecθ) is equal to

(A) 0

(B) 1

(C) 2

(D) -1

Solution:

= (1 + tanθ + secθ) (1 + cotθ – cosecθ)

= (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ)

= (cosθ + sinθ + 1/cosθ) (sinθ + cosθ – 1/sinθ)

= (cosθ + sinθ)2 – (1)2/cosθ sinθ [∵ a2 – b2 = (a + b (a – b)

= cos2θ+sin2θ + 2sinθcosθ-1/cos sinθ

= 1 + 2sinθ cosθ -1/cosθ sinθ

= 2 (C)

(9) a cotθ + b cosecθ = P and b cotθ + a cosecθ = q then p2 – q2 is equal to

(A) a2 – b2

(B) b2 – a2

(C) a2 + b2

(D) b – a

Solution:

p2 – q2

= (a cotθ + bcosecθ)2 – (b cotθ + a cosecθ)2

= (a2cot2θ + 2ab cotθ.cosaθ+cosec2θ) – (b2cot2θ + 2ab cotθcosθ + a2cosec2θ)

= a2 cot2 + 2ab cotθ.cosecθ + b2 cosec2θ – b2 cot2θ – 2abcotθ.cosecθ – a2 cosec2θ

= b2 (cosec2θ – cot2θ) – a2 (cosec2θ – cot2θ)

= b2 – a2 (B)

(10) If the ratio of the height of a tower and the length of its shadow is√3:1, then the angle of elevation of the sun has measure

(A) 45°

(B) 30°

(C) 90°

(D) 60°

Solution: Now, AB:BC = √3:1

AB/BC = √3/1

Let, the angle is θ

∆ABC, tanθ = AB/BC

tanθ = √3/1

tanθ = tan60°

θ = 60° (D)

(11) The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘b’ metres above the first, the depression of the foot of the pole is 60°. The height of the pole (in metres) is equal to

(A) √3 b

(B) b/3

(C) b/2

(D) b/√3

Solution: Now, DC = b meters

∆DCB, tan60° = DC/BC

√3 = b/BC

BC = b/√3

Now, ∆ABC

tan30° = AB/BC

1/√3 = AB/(b/√3)

AB = b/√3×√3 = b/3

∴The height of the pole is b/√3 meters (B)

(12) A tower is 60m height. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30°, then x is equal to

(A) 41.92m

(B) 43.92m

(C) 43m

(D) 45.6m

Solution: Now, AB = tower height = 60m

CD = x meters.

Now, ∆ABC,

tan45° = AB/BD

1 = 60/BD

BD = 60m

And ∆ABC,

tan30° = AB/BC

1/√3 = 60/BC

BC = 60√3

BC = 60×1.732 (√3 = 1.732)

BC = 103.92

∴ DC = x = BC – BD

= (103.92 – 60) m

= 43.92 m. (B)

(13) The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in metres) is

(A) 20, 10√3

(B) 30, 5√3

(C) 20, 10

(D) 30, 10√3

Solution: DC = 20m

∆AED, tan30° = AE /ED

1/√3 = x-20/ED

ED = (x – 20) √3

Let, AB = x m.

AE = (x – 20) m

∆ABC,

tan60° = AB/BC

√3 = x/BC

BC = x/√3

Now, ED = BC

(x – 20) √3 = x/√3

3x – 60 = x

2x = 60

x = 30m

∴ BC = 30/√3 = 10√3 m.

∴The height of the multistoried building and the distance between two buildings is 30 and 10√3. (D)

(14) Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining theirfeet an observer finds the angular elevations of their tops to be complementary, thenthe height of the shorter person (in metres) is

(A) √2 x

(B) x/2√2

(C) x/√2

(D) 2x

Solution: BD = x m

Then, BC = x/2 m and CD = x/2 meters.

Let, ∠ACB = θ then, ∠ECD = (90 – θ).

Let, ED = h then AB = 2h.

∆ABC,

tanθ = AB/BC

tanθ = 2h/(x/2)

tanθ = 4h/x —- (i)

And ∆ECD,

tan (90 – θ) = ED/CD

cotθ = h/(x/2)

cotθ = 2h/x

tanθ = x/2h —– (ii)

∴ Now, from (i) and (ii), we get,

4h/x = x/2h

x2 = 8h2

h2 = x2/8

h = √x2/4×2

h = x/2√2 meters

Thus, the height of the shorter person is x/2√2 meters (B)

(15) The angle of elevation of a cloud from a point h meters above a lake is β. The angle of depression of its reflection in the lake is 45°. The height of location of the cloud from the lake is

(A) h(1+tanβ)/1-tanβ

(B)h(1-tanβ)/1+tanβ

(C) h tan (45° – β)

(D) None of these

Solution: Given, BD = h meters

Let, BC = x. DC1 = (x + h) meters

∆EDC,

tanβ = DC/ED

tanβ = (x-h)/ED

ED = x-h/tanβ —- (i)

Now, ∆EDC1,

tan45° = DC1/ED

1 = x+h/ED

ED = x + h

x-h/tanβ = x + h [By equation (i)]

x – h = xtan β + h tanβ

x (1 – tanβ) = h (tanβ + 1)

x = h (1 + tanβ)/(1 – tanβ)

Thus, the height of location of the cloud from the lake is h(1+tanβ)/(1-tanβ) (A)

Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 6 Trigonometry Exercise 6.5

Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.

Updated: March 2, 2022 — 4:43 pm