I have the following triangle:

The side going up has a length of 96, the side going down has a length of 112. The angle closest to the center is 91 degrees broken up into 62 and 29 degrees from the x-axis. I am trying to find the length of the third side. I was trying to use the formula $C^2=A^2+B^2-2AB *cos(c)$, but I seem to get a different answer than I should. I used the longest side as $C$, the side with the length of 96 as $B$, and the side with the length of 112 as $A$. When I use the above formula, the third side equals 148.8, however, If I were to do it by using 2 triangles and a distance formula (done bellow), I end up with 183.03

The second way I did it was as follows:

For the top triangle:

  • $\cos(62)\cdot96=45.1$
  • $\sin(62)\cdot96=84.8$
  • Therefor the top points coordinates are (+45.1, +84.8)

For the Bottom triangle:

  • $\cos(29)\cdot112=98.0$
  • $\sin(29)\cdot112=54.3$
  • Therefor the bottom point's coordinates are (+54.3, -98.0)

Then I just plugged it into $A^2+B^2=C^2$


Which is totally different than:


What am I doing wrong? Is the formula incorrect?

  • 1
    $\begingroup$ It looks like the bottom point has coordinates $(98, -54.3)$. The Law of Cosines computation looks correct. $\endgroup$
    – Patrick
    Mar 27 '12 at 23:13
  • $\begingroup$ You may wish to keep in mind that if $\alpha$, $\beta$, and $\gamma$ are angles in a triangle, then $\alpha+\beta+\gamma=180^{\circ}$. This should make your trigonometric career much, much easier. $\endgroup$
    – 000
    Mar 28 '12 at 1:20

The second point has coordinates $(98, -54.3)$.


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.