SOLUTION: If A and B are acute angles such that sin(A) = 3/5 and cos(B) = 5/13, find tan(A+B) without evaluating A or B

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Question 1108772: If A and B are acute angles such that sin(A) = 3/5 and cos(B) = 5/13, find tan(A+B) without evaluating A or B
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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1.  If  sin(A) = 3%2F5,  then  cos(A) = sqrt%281-sin%5E2%28A%29%29 = sqrt%281-%283%2F5%29%5E2%29 = sqrt%281-9%2F25%29 = sqrt%28%2825-9%29%2F25%29 = sqrt%2816%2F25%29 = 4%2F5,  and, hence,  tan(A) = sin%28A%29%2Fcos%28A%29 = %28%283%2F5%29%29%2F%28%284%2F5%29%29 = 3%2F4.


2.  If  cos(B) = 5%2F13,  then  sin(B) = sqrt%281-cos%5E2%28B%29%29 = sqrt%281-%285%2F13%29%5E2%29 = sqrt%281-25%2F169%29 = sqrt%28%28169-25%29%2F169%29 = sqrt%28144%2F169%29 = 12%2F13,  and, hence,  tan(B) = sin%28B%29%2Fcos%28B%29 = %28%2812%2F13%29%29%2F%28%285%2F13%29%29 = 12%2F5.


3.  Now  tan(A+B) = %28tan%28A%29+%2B+tan%28B%29%29%2F%281+-+tan%28A%29%2Atan%28B%29%29 = %283%2F4%2B12%2F5%29%2F%281-%283%2F4%29%2A%2812%2F5%29%29 = %28%2815%2B48%29%2F20%29%2F%28%281-36%2F20%29%29 = %28%2863%2F20%29%29%2F%28%281-36%2F20%29%29 = %28%2863%2F20%29%29%2F%28%28-16%2F20%29%29 = -63%2F16.

Answer.   tan(A+B) = -63%2F16.