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 $ABC$. picture of triangle

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?

  • 3
    $\begingroup$ no. part (a) is the Law of Cosines. After that, the Law of Sines gives you the other angles $\endgroup$ – Will Jagy Aug 6 '18 at 18:07
  • $\begingroup$ . . . such as this. Pythagoras applies to right triangles. Plenty of other sources are available. $\endgroup$ – Weather Vane Aug 6 '18 at 18:11
  • $\begingroup$ Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You cannot apply the Pythagorean Theorem here since you do not have a right triangle. $\endgroup$ – N. F. Taussig Aug 6 '18 at 18:12

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

$=\displaystyle 12.5235$

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

$=\displaystyle 28.16°$

You can check the answers here: enter link description here

enter link description here


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

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

  • $\begingroup$ I got C=28.16°. Does this seem correct? $\endgroup$ – Ella Aug 6 '18 at 19:52
  • 1
    $\begingroup$ $C=28.15°$ would be slightly better, because it is not affected by the rounding to a three digit number of $b$. $\endgroup$ – random Aug 6 '18 at 20:21
  • $\begingroup$ Here's the cosine law: google.ca/… $\endgroup$ – mvr950 Aug 6 '18 at 20:28

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.