Question:
If sin 3θ = cos (θ − 6°), where (3θ) and (θ − 6°) are both acute angles, then the value of θ is
(a) 18°
(b) 24°
(c) 36°
(d) 30°
Solution:
Given that,
$\sin 3 \theta=\cos (\theta-6)$ …… (1)
We have to find θ
Here the angles (3θ) and (θ − 6) are acute angles and we know that
Therefore we can rewrite the equation (1) as
$\cos (90-3 \theta)=\cos (\theta-6)$
$\Rightarrow \quad 90-3 \theta=\theta-6$
$\Rightarrow \quad 4 \theta=96$
$\Rightarrow \quad \theta=\frac{96}{4}=24^{\circ}$
Therefore option $(b)$ is correct.