Question

If sinA =3/5 and A+B =90 find cosB

Answer 1

$A+B=90^o\to B=90^o-A\\\\\cos B=\cos(90^o-A)=\cos90^o\cos A+\sin90^o\sin A\\\\=\cos A+1\cdot\dfrac{3}{5}=0+\dfrac{3}{5}=\dfrac{3}{5}\\\\\boxed{\cos B=\dfrac{3}{5}}$

$Used:\\\\\cos(x+y)=\cos x\cos y+\sin x\sin y\\\\\cos90^o=0,\ sin90^o=1$

Other method:

Look at the picture.

$\sin\theta=\dfrac{opposite}{hypotenuse}\\\\\sin A=\dfrac{3}{5}\to opposite=3\ and\ hypotenuse=5$

$\cos\theta=\dfrac{adjacent}{hypotenuse}\\\\adjacent=3\ and\ hypotenuse=5\to\cos B=\dfrac{3}{5}$