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If tan (A + B) = √3 and tan (A – B) = \(\frac{1}{\sqrt 3}\) 0° < A + B ≤ 90°; A > B. find A and B. |
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Answer» tan (A + B) = √3 ⇒ A + B = 60° ………….. (i) tan (A – B) = \(\frac{1}{\sqrt3}\) ⇒ A – B = 30° ………… (ii) From Eqn. (i) + Eqn.(ii), we have 2A = 90 ⇒ = 45° From eqn. (i), A + B = 60° 45° + B = 60° ∴ B = 15 |
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