Finding the side and angle of a triangle.

Question

I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.

The following diagram shows the triangle $ABC$.

a. Find $AC$.

b. Find $\angle BCA$.

For a, I believe I would do the Pythagorean Theorem to find the side. $a^2 + b^2 = c^2$. Is this correct?

For b, to find this angle would I use the sides? As in using soh-cah-toa? So, I could do the sine of $6$ over the hypotenuse, which I would find after part a.

Edit: After reading comments, I used the Law of Cosines for part a and got b = 12.5 as my answer. However, I am not sure I completed it correctly.
I also used the Law of Sines to do part b, I got C = 28.16° as my answer. Can someone please tell me if I completed these two correctly?

Answer 1

$\displaystyle \sqrt{(36 + 100 - ((2 * 6 * 10)* \cos{(100)}))}$

$=\displaystyle 12.5235$

$\displaystyle \sin^{-1} \left(\frac{6 * \sin(100)}{12.5235} \right)$

$=\displaystyle 28.16°$

You can check the answers here: enter link description here

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Answer 2

$\text{ We can also do this. We know through sine law: }$ $\displaystyle \frac{AC}{\sin(\angle{ABC})} = \frac{AB}{\sin(\angle{BCA})}$

$\displaystyle \angle{BCA} = \sin^{-1}\left(\frac{AB * \sin(\angle{ABC})}{AC}\right)$

Answer 3

For $a$ we can't use Pythagorean Theorem since ABC is not a right triangle but we need the Law of cosines.

For that see the related Does the law of cosines contradict Pythagoras's theorem?

For point b once we have AC by the Law of sines we have

$$\frac{\sin 100}{AC}=\frac{\sin (\angle BCA)}{AB}$$