Question

In `triangle MNP, m=16, n=15, p=8`, how do you find the cosine of each of the angles?

Answer 1

`cos hatM=0.1375`
`cos hatN=0.37109375`
`cos hatP=0,86875`

Explanation:

If you know two sides and the included angle of a triangle ABC, you can apply the cosine thorem:

`a^2=b^2+c^2-2bc cos hatA`

Now you can use the inverse formula to know the cosine of an angle when you know the three sides:

`cos hatA=(b^2+c^2-a^2)/(2bc)`

Then

`cos hatM=(n^2+p^2-m^2)/(2np)`

`=(15^2+8^2-16^2)/(2*15*8)`

`=0.1375`

`hatM=82.1°`

`cos hatN=(m^2+p^2-n^2)/(2mp)`

`=(16^2+8^2-15^2)/(2*16*8)`

`=0.37109375`

`hatN=68.22°`

`cos hatP=(m^2+n^2-p^2)/(2mn)`

`=(16^2+15^2-8^2)/(2*16*15)`

`=0,86875`

`hatP=29.69°`