Question

If sin 3a = cos (a-6) where 3a and a-6 are both

Answer 1

I think question is like this ,

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°

Additional formuee--->

(1) Cos ( 90° - θ ) = Sinθ

(2) tan ( 90° - θ ) = Cotθ

(3) Cot ( 90° - θ ) = tanθ

(4) Sec ( 90° - θ ) = Cosecθ

(5) Cosec( 90° - θ ) = Secθ

Answer 2

Answer:

Step-by-step explanation:

sin 3A = cos (A-6)

sin 3A = sin (90-(A-6))

( since sin(90- A) = cos A)

therefore,

3A = 90-(A-6)

3A = 90-A+6

3A+A = 96

4A = 96

A = 96/4

A = 24