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

Answer:

$sin(A+B) = \frac{1431}{1769}$

Step-by-step explanation:

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

$given : cos A = \frac{20}{29}, \ sinB = \frac{60}{61}$

First we will find sinA and cosB

$We \ know \ sin^2 A + cos^2A = 1\\\\sin^2 A = 1 - cos^2A = 1 - (\frac{20}{29})^{2} = \frac{841-440}{841} = \frac{441}{841}\\\\sinA =\frac{21}{29}$

$sin^2B +cos^2B = 1\\\\cos^2B = 1 - sin^2B = 1 - (\frac{60}{61})^2 = \frac{3721-3600}{3721} = \frac{121}{3721}\\\\cosB = \frac{11}{61}$

Substitute all values in (1)

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

               $=(\frac{21}{29} \times \frac{11}{61} )+ (\frac{20}{29}\times \frac{60}{61})\\\\=\frac{231}{1769} + \frac{1200}{1769}\\\\= \frac{1431}{1769}$