Question

For positive acute angles AA and B,B, it is known that \cos A=\frac{24}{25}cosA= 25 24 ​ and \sin B=\frac{11}{61}.SinB= 61 11 ​ . Find the value of \sin(A+B)sin(A+B) in simplest form.

Answer 1

Answer:

684/1525

Step-by-step explanation:

Given

cosA = 24/25

SinB = 11/61

Sin(A+B) = SinAcosB + CosAsinB

Get SinA

Since

Cos A = adj/hyp

adj = 24

hyp = 25

opp² = 25² - 24²

opp² = 625 - 576

opp² = 49

opp = √49

opp = 7

sinA = opp/hyp

sinA = 7/25

Also if SinB = 11/61

opposite = 11

Hypotenuse = 61

adj² = 61² - 11²

adj² = 3,721 - 121

adj² = 3600

adj = √3600

adj = 60

Cos B = adj/hyp

CosB = 60/61

Sin(A+B) = SinAcosB + CosAsinB

Sin(A+B) = 7/25(60/61) + 24/25(11/61)

Sin(A+B) = 420/1,525 + 264/1,525

Sin(A+B) = 684/1525