If tan A = 5/12 and sin B = 40/41 and angles A and B are in Quadrant I, find the value of tan(A+B).

Question

Help me, please! It's due by 1:30!!!

Answer 1

these are 5-12-13 and 9-40-41 right triangles. sinA = 5/13, cosA =12/13, cosB = 9/41, sinB = 40/41

tan(A+B) = sin(A+B)/cos(A+B) = (sinAcosB + sinBcosA)/cosAcosB - sinAsinB) plug in the values

(5/13)(9/41) +(40/41)(12/13)/(12/13)(9/41)-(5/12)(40/41) = (45/533 + 480/533)/(108/533 - 200/533) =

525/-92

or

tan(A+B) = (tanA + tanB)/(1 - tanAtanB) = (5/12 + 40/9)/(1-(5/12)(40/9)= 525/-92