Question

URGENT‼️‼️‼️ solve the triangle using law of sines and cosines.

Answer 1

`m\angle C=20^(\circ)\\m\angle B = 127^(\circ)\\b=37\\`

Explanation:

The Law of Sines for any triangle is:

`(\sin A)/(a)=(\sin B)/(b)=(\sin C)/(c)`

Let's solve for
`m\angle C` first:

`(\sin33^(\circ))/(25)=(\sin (m\angle C))/(16),\\\sin (m\angle C)=(16\cdot \sin33^(\circ))/(25), \\m\angle C= \sin^(-1)((16\cdot \sin33^(\circ))/(25))\approx \fbox{$20^(\circ)$}`

Since the sum of the interior angles in a triangle is always
`180^(\circ)`, the measure of angle B is
`m\angle B=180-33-20=\fbox{$127^(\circ)$}`.

Now we can use to solve for side length
`b`:

`(\sin33^(\circ))/(25)=(\sin 127^(\circ))/(b),\\b=(25\cdot \sin127^(\circ))/(\sin33^(\circ))\approx \fbox{37}`.