Question

In Triangle HIJ, if h=10, j=7, and cosI=0.6, how do you find the exact value of i?

Answer 1

`i=8.0623`

Explanation:

To find `i`, we need to use cosine formula for triangles.

As per this, in a `DeltaHIJ`,

`i^2=h^2+j^2-2xxhxxjxxcosI`

= `10^2+7^2-2xx10xx7xx0.6`

= `100+49-84`

= `65`

Hence `i=sqrt65=8.0623`