$C^2=A^2+B^2-2AB \cdot\cos(c)$ getting a different answer than creating a third triangle with the distance formula?

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 $A^2+B^2=C^2$

$\sqrt{(54.3-45.1)^2+(-98-84.8)^2}=\sqrt{(9.2)^2+(-182.8)^2}=183.03$

Which is totally different than:

$\sqrt{(96)^2+(112)^2-2(96*112)\cdot\cos(91)}=148.8$

What am I doing wrong? Is the formula incorrect?

• It looks like the bottom point has coordinates $(98, -54.3)$. The Law of Cosines computation looks correct. Mar 27 '12 at 23:13
• 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.
– 000
Mar 28 '12 at 1:20

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