1.

If sin3A =  cos(A -10°) and 3A is acute then ∠A = ?(a) 35° (b) 25° (c) 20°(d) 45°

Answer»

Correct answer = (b) 25°

We have

[sin3A =  cos(A -10°)]

=> cos(90° − 3A) = cos(A − 10°) [∵ sin θ = cos(90° − θ)] 

=> 90° − 3A = A − 10° 

=> −4A = −100 

=> A = 100/4 

=> A = 25°



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