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