If sin 3θ = cos (θ − 6°),


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°


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.


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