Question

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

Answer 1

Answer:

Solution: tan (A + B) = √3 ⇒ tan (A + B) = tan 60° ⇒ (A + B) = 60° ... (i) tan (A – B) = 1/√3 ⇒ tan (A - B) = tan 30° ⇒ (A - B) = 30° ... (ii) Adding (i) and (ii), we get A + B + A - B = 60° + 30° 2A = 90° A= 45° Putting the value of A in equation (i) 45° + B = 60° ⇒ B = 60° - 45° ⇒ B = 15° Thus, A = 45° and B = 15°

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Answer 2

Answer:

A= 45° B= 15°

Step-by-step explanation:

Given tan(A+B)=√3⟹A+B=60∘⋯(1)

tan(A−B)=31⟹A−B=30∘⋯(2)

(1)+(2)⟹2A=90∘⟹A=45∘

Put A value in (1)

⟹B=60∘−45∘=15∘