Question

If sin 3A = cos (A – 10°), where 3A is an acute angle then find ∠A.

Answer 1

Given: sin 3A = cos (A – 10°) and 3A is an acute angle.

Step-wise calculation:

1. cos(A – 10°) = sin[90° – (A – 10°)]

= sin(100° – A)

So sin 3A = sin(100° – A).

2. Therefore either (i) 3A = 100° – A + 360k, or

(ii) 3A = 180° – (100° – A) + 360k

= 80° + A + 360k, for some integer k.

3. For k = 0: (i) 3A = 100° – A

⇒ 4A = 100° 

⇒ A = 25°

Then 3A = 75°, which is acute, acceptable.

(ii) 3A = 80° + A

⇒ 2A = 80° 

⇒ A = 40°

Then 3A = 120°, not acute, reject.

4. Other integer k give A outside the range consistent with 3A being acute, so no further solutions.

∠A = 25°