1.

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

Answer»

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°



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