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(3)/(5)=0+(3)/(5)=(3)/(5)\\\\\boxed{\cos B=(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=(opposite)/(hypotenuse)\\\\\sin A=(3)/(5)\to opposite=3\ and\ hypotenuse=5`

`\cos\theta=(adjacent)/(hypotenuse)\\\\adjacent=3\ and\ hypotenuse=5\to\cos B=(3)/(5)`