How to use the Law of Sines to Find an Angle

Question

I am trying to figure out how to find an angle with the law of sines.

I have a triangle where:

A = $120^\circ$

B = unmarked

C = $\theta$

a = 45

b = unmarked

c = 36

How can I find the angle for C?

I have tried:

$$\frac{sin120^\circ}{45} = \frac{sinB}{b} =\frac{sin\theta}{36} $$

$$36(\frac{sin\theta}{36}) = 36(\frac{sin120^\circ}{45})$$ $$sin\theta = \frac{36sin120^\circ}{45}$$ $$ \theta = \frac{36(120^\circ)}{45} = 95^\circ$$

But, the answer is supposed to be $44^\circ$.

Answer 1

The following step is completely wrong

$$\sin\theta = \frac{36 \sin120^\circ}{45} \iff \color{red}{\theta = \frac{36(120^\circ)}{45} = 95^\circ}$$

we have that $\sin120^\circ=\sqrt 3/2$ and then

$$\sin\theta = \frac{36\sin120^\circ}{45} \iff \sin \theta = \frac{36\sqrt 3}{2\cdot 45} \iff \theta = \arcsin \left(\frac{2\sqrt 3}{9}\right)$$

Answer 2

Most of your steps are correct, but it is from the second last step you made a fundamental mistake.

Using arcsine ($sine^{-1}$ on calculator):

$$sin\theta = \frac{36sin120^\circ}{45} $$ $$\theta = arcsine(\frac{36sin120^\circ}{45}) $$ $$\theta = 43.8537^{\circ}... $$ $$\theta \approx 44^{\circ} $$

Just as @Dr. Sonnhard Graubner pointed out. The mistake you made was that you interpreted arcsine and sin$\theta$ canceling out, because sometimes it is learned to students that $arcsine = \frac{1}{sin}$. The thing is, $sin$ on itself cannot be cancelled out, as it is a trigonometric function of an angle, not $s . i . n$ as variables.