Q12 of 268 Page 4

If sin 3A = cos(A – 26^o), where 3A is an acute angle, find the value of A.

sin 3A = cos (A-26°) …(i)

We know that

Sin θ = cos (90° - θ)

So, Eq. (i) become

Cos (90° - 3A) = cos (A -26°)

On Equating both the sides, we get

90° - 3A = A – 26°

⇒ -3A - A = -26° -90°

⇒ -4A = -116°

⇒ A = 29°

In a ∆ABC prove that

Find θ if cos(2 θ +54^o)= sin θ, where (2θ +54^o) is an acute angle.

If tan 3 θ =cot (θ +18^o), where 3 θ and θ +18^o are acute angles, find the value of θ.

If sin 3A = cos(A – 26o), where 3A is an acute angle, find the value of A.