Question

For an Acute angle sin 3A° = cos (A – 6)°. Fine the value of A.

Answer 1

Sin3A = Cos ( A - 6 ) , where 3A and A - 6 are both acute angle then find value of A.

Solution---> ATQ,

Sin3A = Cos ( A - 6 )

We have a formula

Cosθ = Sin ( 90° - θ ) , applying it we get,

=> Sin3A = Sin { 90° - ( A - 6 ) }

=> Sin3A = Sin ( 90° - A + 6 )

=> 3A = 90° - A + 6

=> 3A + A = 96

=> 4A = 96

=> A = 96 / 4

=> A = 24°

Answer 2

Answer:

The value of A is 24°.

Step-by-step explanation:

Given :

➸ sin 3A° = cos (A – 6)°

➸ Both are acute angles

We know that :

➸ sin θ = cos (90 – θ)

So :

➸ sin 3A = cos (A – 6)

➸ cos (90 – 3A) = cos (A – 6)

Let's solve for "A" :

➸ 90 – 3A = A – 6

➸ 90 + 6 = A + 3A

➸ 96 = 4A

➸ 96/4 = A

➸ A = 24°

∴ The value of "A" is 24°.

Verification:

Let's substitute by taking A = 24° :

➸ sin 3A° = cos (A – 6)°

➸ sin 3 (24°) = cos (24° – 6)°

➸ sin 72° = cos 18°

➸ √(10+2√5/4) = √(10 + 2√5/4)

➸ LHS = RHS