Trigonometry

[ResidentBiscuit]

Here's another problem:

The sides of a parallelogram are 14 ft and 16 ft. One angle is 48° while another angle is 132°. Find the lengths of the diagonals of the parallelogram (to the nearest tenth of a foot).

I got 152.2 and 751.8, which seem wrong to me. Not sure what I could have done wrong though :/

[ne555]

¿seem wrong? you don't respect the triangle.
show some code and we may help you, but don't expect that people will do your homework.

[ResidentBiscuit]

It's not a program lol. It's math homework, hence the title of the thread. Not asking for anyone to do my homework.

[ne555]

I know, but the same rules apply.
You asked what is wrong, but didn't show what you did.

[ResidentBiscuit]

I was going to, but as I can't really draw on here it's an issue. I'll give a shot though:

I cut this into two triangles, one with the long side going towards the 48° angles, and the other's long side going towards the 132° angles.

This created two triangles with angles:

24 132 24
AND
48 66 48

I knew two sides for each triangle, so I plugged them into the law of cosines equation:

a^2 = b^2 + c^2 - 2bc cos(A)

From there, I just solved and came out the numbers above. According the triangle side length restriction, this can't be true because I have triangles with side lengths of:
14 16 751
AND
14 16 152

14 + 16 < 152 < 751

EDIT:
Just realized a pretty silly mistake. 48 66 48 isn't even a triangle. 48 + 66 + 44 != 180. Doesn't explain the other triangle though

[ne555]

You forgot to compute the square root.

Edit: ¿who told you that the triangle were isosceles?

[ResidentBiscuit]

Wow! I figured I missed something small there. That explains the numbers not adding up.

Anyway, thanks ne555!