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If tan (A − B) = 1/√3 and tan (A + B) = √3, 0° < (A + B) < 90° and A > B, then find A and B. |
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Answer» tan (A – B)= 1/√3 or tan(A – B) = tan 30° A – B = 30° …(1) Again, tan(A+B) = √3 = tan 60° A + B = 60° …(2) Solving (1) and (2), we get 2A = 90° or A = 45° Putting A = 45° in (1), we get 45° – B = 30° or B = 45° – 30° = 15° Therefore, A = 45°, B = 15° |
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