Question

Determine mzB if a = 14, b = 15.0 = 17 of AABC. Determine theangle to the nearest tenth of a degree.(Here A is the angle opposite side a. B is the angle opposite side b, etc.)

Answer 1

Given:

a=14

b=15

c=17

Find measure of angle B.

Applying law of cosines, we have:

$b^2=a^2+c^2-2ac\cos (B)$

Putting the values,we have:

$15^2=14^2+17^2-2\times14\times17\times\cos (B)$

We have to find B so ,

$\cos \text{ (B)=}\frac{196+289-225}{476}$$\begin{gathered} \cos (B)=\frac{260}{476} \\ =0.5462 \end{gathered}$$B=\cos ^{-1}\text{ 0.5462}$

Hence, the angle B is nearly 60 degrees.

That is,

$m\angle B=60^{\circ}$

This is required answer.