Question

Please check!!!! I've posted this already but no one is answering so I'm posting for more points.

Answer 1

By the law of sines,

$\dfrac{\sin m\angle A}a=\dfrac{\sin m\angle B}b\implies\sin m\angle B=\dfrac{33.7\sin75^\circ}{51.2}$

We get one solution by taking the inverse sine:

$m\angle B=\sin^{-1}\dfrac{33.7\sin75^\circ}{51.2}\approx39^\circ$

In this case there is no other solution!

To check: suppose there was. The other solution is obtained by recalling that $\sin(180-x)^\circ=\sin x^\circ$ for all $x$, so that

$180^\circ-m\angle B=\sin^{-1}\dfrac{33.7\sin75^\circ}{51.2}\implies m\angle B\approx141^\circ$

But remember that the angles in any triangle must sum to 180 degrees in measure. This second "solution" violates this rule, since two of the known angles exceed 180: 75 + 141 = 216 > 180. So you're done.

Answer 2

This triangle is not a right triangle. How do we solve this then? You will use the law of sine with is shown below:

$\frac{sin A}{a} =\frac{sin B}{b} = \frac{sinC}{c}$

What we know is shown in the image attached below:

Plug what you know into the law of sine

$\frac{sin75}{51.2} =\frac{sinB}{33.7}$

To solve for sinB cross multiply

sin75*33.7 = sinB * 51.2

32.55 = sinB*51.2

Divide 51.2 to both sides to isolate sinB

32.55 / 51.2 = sinB / 51.2

0.63577 = sinB

To find B you must use arcsin:

$sin^{-1} 0.63577$

39.477

^^^This is your rough estimate but you can simply keep it to 39 degrees

This means that your answer is correct!

Hope this helped!