For positive acute angles AA and B,B, it is known that \sin A=\frac{3}{5}sinA= 5 3 ​ and \tan B=\frac{9}{40}.tanB= 40 9 ​ . Find the value of \cos(A+B)cos(A+B) in simplest form.

Question

For positive acute angles A and B, it is known that sinA=53​ and tanB=409​. Find the value of cos(A+B) in simplest form.

Answer 1

The formatting of the problem is confusing. After all, there is no sense in which sinA = 53. Sin varies between -1 and +1.

let’s guess that the original problem is

sin A = 3/5

tan B = 9/40

find cos(A+B)

first let’s hope these are both all-integer triangles:

In a 3-4-5 Right triangle,

sinA = 3/5, cos A = 4/5

What about 9,40? It turns out the hypotenuse is 41, so

sin B = 9/41

cos B = 40/41

So now we just use the identity for cos ( A + B )

Cos(A+B) = cosAcosB - sinAsinB

= (4/5)(40/41)-(3/5)(9/41)

=(4*40-3*9) / (5*41)

=(160-27)/(5*41)

= 133/205