If sin 3A = cos (A – 10º) and 3A is acute then \angleA = ?

Foundation

Mathematics Foundation

Introduction to Trigonometry

Question

If sin 3A = cos (A – 10º) and 3A is acute then $∠$ A = ?

35º

20º

45º

25º

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sin 3A = cos (A – 10º) ⇒ cos (90º – 3A) = cos (A – 10º)

⇒ 90º – 3A = A – 10º ⇒ 4A = 100º ⇒ A = 25º.

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