# In triangle ABC, if a = 8.75 centimeters, c = 4.26 centimeters, and m/_B is 87° what is the length of b to two decimal places?

Aug 13, 2018

$\therefore b \approx 9.53 \text{ cm}$

#### Explanation:

We know that ,

$\textcolor{red}{\text{cosine Rule : } \cos B = \frac{{c}^{2} + {a}^{2} - {b}^{2}}{2 c a}}$

$\therefore 2 c a \cos B = {c}^{2} + {a}^{2} - {b}^{2}$

$\therefore {b}^{2} = {c}^{2} + {a}^{2} - 2 c a \cos B \ldots \to \left(1\right)$

We have ,

$a = 8.75 \text{ cm} ,$ , $c = 4.26 \text{ cm} \mathmr{and} B = {87}^{\circ}$

Using $\left(1\right)$ we get

${b}^{2} = {\left(4.26\right)}^{2} + {\left(8.75\right)}^{2} - 2 \left(4.26\right) \left(8.75\right) \cos {87}^{\circ}$

$\therefore {b}^{2} = 18.1476 + 76.5625 - 74.55 \left(0.0523\right)$

$\therefore {b}^{2} \approx 90.81135$

$\therefore b \approx 9.529$ cm