Ex-5.2 Trigonometric Ratios (Part - 2), Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Q26) If θ = 30°,verify(i)Tan2θ = 

Solution:

Tan2θ = …..(i)

Substitute θ = 30° in equation (i)

LHS = Tan 60°  =  √3

RHS =  

Therefore, LHS = RHS

(ii) sinθ =

Substitute θ = 30°

sin60° =

Therefore, LHS = RHS.

(iii) cos2θ =

Substitute θ = 30°

LHS =  cosecθ                       RHS =  

= cos 2(30°)                      =

Cos 60°  =  1/2                  =

Therefore, LHS = RHS

(iv) cos3θ = 4cos³θ − 3cosθ

Solution:

LHS =  cos3θ

Substitute θ = 30°

= cos 3 (30°) = cos 90⁰

= 0

RHS =  4cos³θ − 3cosθ

=  4cos³30° − 3cos30°

= 0

Therefore, LHS = RHS.

Q27) If A = B = 60°. Verify  (i) Cos (A – B) = Cos A Cos B + Sin A Sin B

Solution:

Cos (A – B) = Cos A Cos B + Sin A Sin B…….(i)

Substitute A and B in (i)

⇒ cos (60°  – 60°) = cos 60° cos 60°  + sin 60° sin 60° 

⇒ cos 0⁰  =  (1/2)² + (√3/2)²

⇒ 1 =  1/4 + 3/4

⇒ 1 = 1

Therefore, LHS = RHS

(ii) Substitute A and B in (i)

⇒ sin (60°  – 60°) = sin 60° cos 60°  – cos 60° sin 60°

⇒ sin 0⁰  = 0

⇒ 0 = 0

Therefore, LHS = RHS

(iii) Tan(A − B) = 

A = 60°, B = 60° we get,

Tan(60° − 60°) =

Tan 0⁰  = 0

0 = 0

Therefore, LHS = RHS

Q28 ) If A = 30°, B = 60° verify:

(i) Sin (A + B) = Sin A Cos B + Cos A Sin B

Solution:

A = 30°, B = 60° we get

Sin (30°  + 60°) = Sin 30°  Cos 60°   + Cos 30°  Sin 60°

Sin (90⁰) =  1/2.1/2 + √3/2.√3/2

Sin (90⁰) = 1 ⇒ 1 = 1

Therefore, LHS = RHS

(ii) Cos (A + B) = Cos A Cos B – Sin A Sin B

A = 30°, B = 60° we get

Cos (30°  + 60°) = Cos 30° Cos 60°  – Sin 30° Sin 60°

Cos (90⁰) =  1/2.3√2 − √3/2.1/2

0 = 0

Therefore, LHS = RHS

Q29. If sin(A + B) = 1 and cos(A-B) = 1, 0°<A + B≤90°, A≥B find A and B.

Sol:

Given,

sin(A + B) = 1 this can be written as sin(A + B) =  sin(90°)

cos(A-B) = 1 this can be written as cos(A-B) =  cos(0°)

⇒ A + B =  90°

A – B =  0°

2A       =  90°

A         =  90°/2

A         =  45°

Substitute A value in A – B =  0°

45°  – B =  0°

B = 45°

Hence, the value of A =  45° and B = 45°

Q30. If tan(A-B) =  1/√3 and tan(A + B) =  √3, 0°<A + B≤90°, A>B find A and B

Solution:

Given,

tan(A-B) =  1/√3

A – B   =  tan ^{− 1}(1/√3)

A – B   =  30°                  ——- 1

tan(A + B) =  √3

A + B =  tan ^{− 1}√3

A + B  =  60°                   ——- 2

Solve equations 1 and 2

A + B =  30°

A – B =  60°

2A  =  90°

A =  90°/2

A =  45°

Substitute the value of A in equation 1

45°  + B =   30°

B =  30°  –  45°

B =  15°

The value of A =  45° and B =  15°

Q31. If sin(A-B) =  1/2 and cos(A + B) =  1/2, 0°<A + B≤90°, A<B find A and B.

Solution:

Given,

sin(A-B) =  1/2

A – B =  sin ^{− 1}(1/2)

A – B =  30°                    ——- 1

cos(A + B) =  1/2

A + B =  cos ^{− 1}(1/2)

A + B =  60°                 ——- 2

Solve equations 1 and 2

A + B =  60°

A – B =  30°

2A  =  90°

A =  90°/2

A =  45°

Substitute the value of A in equation 2

45°  + B =   60°

B =  60°  –  45°

B =  15°

The value of A =  45° and B =  15°

Q32. In a Δ ABC right angled triangle at B, ∠A = ∠C.Find the values of:

1. sinAcosC + cosAsinC

Solution:

since, it is given as ∠A = ∠C

the value of A and C is 45°, the value of angle  B is 90°

because the sum of angles of triangle is 180°

⇒ sin(45°)cos(45°) + cos(45°)sin(45°)

⇒  1/2  +  1/2

⇒ 1

The value of  sinAcosC + cosAsinC is 1

2. sinAsinB + cosAcosB

Solution:

since, it is given as ∠A = ∠C

the value of A and C is 45°, the value of angle  B is 90°

because the sum of angles of triangle is 180°

⇒ sin(45°)sin(90°) + cos(45°)sin(90°)

⇒  1/√2(1) +  1/√2(0)

⇒  1/√2  + 0

⇒  1/√2

The value of sinAsinB + cosAcosB is 1/√2

Q33. Find the acute angle A and B, if sin(A + 2B) =  √3/2 and cos(A + 4B) = 0, A>B.

Solution:

Given,

sin(A + 2B) = √3/2

A + 2B =  sin ^{− 1}√3/2

A + 2B =  60°                     —— – 1

Cos(A + 4B) = 0

A + 4B =  sin ^{− 1}(90)

A + 4B =  90°                     —— – 2

Solve equations 1 and 2

A + 2B =  60°

A + 4B =  90°

(-)  (-)         (-)

-2B = – 30°

2B =  30°

B =  30°/2

B =  15°

Substitute B value in eq 2

A + 4B =  90°

A + 4(15°) =  90°

A +  60°  =  90°

A =  90°  –  60°

A =  30°

The value of A =  30° and B =  15°

Q 34. In ΔPQR, right angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠ P and ∠ R.

Solution:

Given,

In ΔPQR, right angled at Q, PQ = 3 cm and PR = 6 cm

By Pythagoras theorem,

PR² = PQ² + QR² 

⇒ 6² = 3² + QR² 

⇒ QR² = 36 – 9

sin R =   3/6 = 1/2 = sin30°

∠R = 30°

As we know, Sum of angles in a triangle = 180

∠P + ∠Q + ∠R = 180°

⇒ ∠P + 90° + 30° = 180°

⇒ ∠P = 180° – 120°

⇒ ∠P = 60°

Therefore, ∠R = 30°

And, ∠P = 60°

Q35. If sin(A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15 and cos 15.

Solution:

Given,

sin(A – B) = sin A cos B – cos A sin B

And, cos (A – B) = cos A cos B + sin A sin B

We need to find, sin 15 and cos 15.

Let A = 45 and B = 30

sin 15 = sin (45- 30) = sin 45 cos 30 – cos 45 sin 30

cos 15 = cos (45- 30) = cos 45 cos 30 – sin 45 sin 30

Q36. In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15 units. Find the remaining angles and sides. 

Solution:

sin60° = x/15

cos60° = x/15

1/2 = x/15

x = 15/2

x = 7.5units

 Q37. In ΔABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find the remaining angles and sides.

Solution:

Here, ∠C = 90° and ∠A = 45°

We know that,

∠A + ∠B + ∠C  =  180°

⇒  45°  +  90°  +  ∠C  =  180°

⇒   135°  +  ∠C  =  180°

⇒  ∠C  =  180°  –  135°

⇒  ∠C  =  45°

The value of the remaining angle C is 45°

Now, we need to find the sides x and y

here,

cos(45) =  BC/AB

1/√2  =  7/y

y =  7√2 units

sin(45) =  AC/AB

x = 7 units

the value of x = 7 units and y =  7√2 units

Q 38 . In a rectangle ABCD , AB = 20 cm , ∠BAC =  60° , calculate side BC and diagonals AC and BD .

Solution:

Let AC = x cm and CB = y cm

Since , cosθ  =  base/hypotenuse

Therefore , cos60° = 20/x

⇒1/2 = 20/x                           [since,cos60° = 1/2]

⇒ x = 40 cm = AC

Similarly BD = 40 cm

Now ,

Since , sinθ  =  perpendicular/hypotenuse

Therefore , sin60° = BC/AC

⇒y = 20√3 cm .

Q39:If A & B are acute angles such that tanA = 1/2 tanB = 1/3 and tan(A + B) =  find A + B.

Solution:

Tan(A + B) =

Tan(A + B) = 5/6 × 6/5

(A + B) = Tan ^{− 1}(1)

(A + B) = 45°

Q 40: Prove that : (√3 − 1)(3 − cot30°) = tan³60 − 2sin60°

Ans:

L.H.S ⇒  (√3 + 1)(3 − cot30°)

= (√3 + 1)(3 − √3)       ∵ cot30° = √3

= (√3 + 1)(√3 − 1)√3

=  ((√3)² − (1)²)√3

=  2√3

R.H.S ⇒  tan36° − 2sin60°

=  (√3)³ − 2×√3/2

=  3√3 − √3

=  2√3

L.H.S = R.H.S

Hence proved.