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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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