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Ex 8.2, 3 If tan (A + B) = √3 and tan (A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B. Given that tan (A + B) = √𝟑 But we know that tan 60° = √3 Thus, tan (A + B) = tan 60° ∴ A + B = 60° tan (A – B) = 𝟏/√𝟑 But we know that tan 30° = 1/√3 Thus tan (A – B) = tan 30° ∴ A – B = 30° Our equations are A + B = 60° …(1) A – B = 30° …(2) Adding (1) and (2) A + B + A – B = 60° + 30° 2A = 90° A = (90°)/2 A = 45° Putting A = "45°" in (1) A + B = 60° 45° + B = 60° B = 60° − 45° B = 15° Hence, A = 45° , B = 15°

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo