Question

for positive acute angles A and B, it is known that cos A=20/29 and sin B=60/61. Find the value of sin(A+B) in simplest form.

Answer 1

`sin(A+B) = (1431)/(1769)`

Explanation:

sin(A+B) = sinA cosB + cosA sinB ------(1)

`given : cos A = (20)/(29), \ sinB = (60)/(61)`

First we will find sinA and cosB

`We \ know \ sin^2 A + cos^2A = 1\\\\sin^2 A = 1 - cos^2A = 1 - ((20)/(29))^(2) = (841-440)/(841) = (441)/(841)\\\\sinA =(21)/(29)`

`sin^2B +cos^2B = 1\\\\cos^2B = 1 - sin^2B = 1 - ((60)/(61))^2 = (3721-3600)/(3721) = (121)/(3721)\\\\cosB = (11)/(61)`

Substitute all values in (1)

`sin(A+B) =sinAcosB + cosAsinB`

`=((21)/(29) * (11)/(61) )+ ((20)/(29)* (60)/(61))\\\\=(231)/(1769) + (1200)/(1769)\\\\= (1431)/(1769)`