cosA=15/17

sinA=?

AC²=AB²+BC²

(17)²=AB²+(15)²

289=AB²+225

289-225=AB²

64=AB²

√64=AB²

8=AB

sinA=AB/AC

sinA=8/17

cos(A) = 15/17. sin^2(A) = 1-cos^2(A) = 1 - (15/17)^2 = 1 - (225/289) = (289-225)/289 = 64/289 . Then

sin(A) = rt(64/289) = (8/17).

8^2 + 15^2 = 17^2

so (8, 15, 17) is a Pythagorean Triple, i.e. sides of a right triangle.

Use that.

Angle A could be an angle where the adjacent side is 15 and the hypotenuse is 17, because cosA = adj/hyp.

The using the Pythagorean Theorem, the opposite side must be √17^2 - 15^2 = √289 -225 = √64 = 8

So sinA = 8/17.

You can also use the identity, which you should know, that cos^2 A + sin^2 A = 1.

The arithmetic works out exactly the same way:

sin A = √1 - cos^2 A

= √1 - 15^2/17^2 = √(289 - 255)/289 = √64/289 = 8/17.

8/17