Q15 of 48 Page 86

If sin 3θ = cos (θ – 6°) where 3θ and (θ – 6°) are acute angles then find the value of θ.

As we know,


cos θ = sin (90˚ – θ)


⇒ sin 3θ = cos (θ – 6˚)


⇒ sin 3θ = sin (90˚ – (θ – 6˚))


⇒ sin 3θ = sin (90˚ – θ + 6˚)


⇒ sin 3θ = sin (96˚ – θ)


Since, 3θ is acute angle


So, 3θ = 96˚ – θ


⇒ 4θ = 96˚


⇒ θ = = 24˚


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