# Math Help - Parallelogram Angle Solving

1. ## Parallelogram Angle Solving

Basically just find this angle,

I used the sine rule and got something like 50.8...etc, And when I rounded it off I got 51.
But I dont think its right, Anyone help, I think the red 29 that I've indicated is right because its a parallelogram but if its not then I know I went wrong, Any help on how to solve this.
Thanks

edit: calculated again, I think the right answer was 50*51

Basically just find this angle,

I used the sin rule and got something like 50.8...etc, And when I rounded it off I got 51.
But I dont think its right, Anyone help, I think the red 29 that I've indicated is right because its a parallelogram but if its not then I know I went wrong, Any help on how to solve this.
Thanks

Why don't you think it's right? You're drawing is not drawn to scale, but, other than that, it looks fine.

Lean your parallelogram the other way and the measures make sense.

The problem is the dreaded "ambiguous case".
There are two solutions.

Find angle $x.$
Code:
          C o - - - - - - - - - - - o D
\  *                    \
\  x  *                 \
\        *              \
\ 5         *           \
\              *        \
\                 *     \
\               29°  *  \
A o - - - - - - - - - - - o B
8

Law of Sines: . $\frac{\sin x}{8} \:=\:\frac{\sin29^o}{5} \quad\Rightarrow\quad \sin x \:=\:\frac{8\sin29^o}{5} \:=\: 0.775695392$

Therefore: . $x \:\approx\:\boxed{50.9^o}$

But inverse sine can have more than one value.

Hence: . $x \:=\:180^o - 50.9^o \;=\;\boxed{129.1^o}$

And the parallelogram looks like this:

Code:
          C * - - - - - - - - - - - - - - - - - - - * D
/    *                                  /
/ 129.1°  *                             /
/              *                        /
5 /                   *                   /
/                        *              /
/                             *         /
/                             29°  *    /
A * - - - - - - - - - - - - - - - - - - - * B
8