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

Jan 6, 2017

$i = 8.0623$

#### Explanation:

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

As per this, in a $\Delta H I J$,

${i}^{2} = {h}^{2} + {j}^{2} - 2 \times h \times j \times \cos I$

= ${10}^{2} + {7}^{2} - 2 \times 10 \times 7 \times 0.6$

= $100 + 49 - 84$

= $65$

Hence $i = \sqrt{65} = 8.0623$