Question

In triangle abc , what is the value of cos B?

Answer 1

Sin a = cos b

sin a = sin (90-b)

a =90-b

a+b =90,

From angle sum property, we know that

In a triangle, a + b + c =180

90+c =180

c =90°.

Now cos90° = 0.

Therefore, cosc =0

Hope this helps!