Question 3:

If tan (A + B) = √3 and tan (A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

Answer:

tan (A + B) = √3

⇒ tan (A + B) = tan 60

⇒ A + B = 60 …..(1)

tan (A – B) = 1/√3

⇒ tan (A – B) = tan 30

⇒ A – B = 30 ….. (2)

On adding both equations, we obtain

2A = 90

⇒ A = 45

From equation (1), we obtain

45 + B = 60

B = 15

Therefore, ∠A = 45° and ∠B = 15°

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