# Solving A Parallelogram

If the length of larger side of a parallelogram is $55cm$ and the one diagonal of the parallelogram makes angle of measures $30^\circ$ and $50^\circ$ with of a pair of adjacent sides find the length of the diagonal.

At least give me a base concept or start so I can solve it by my own :)

~Thanks!

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Try drawing a picture. – Daryl Dec 31 '12 at 8:46

1. Use the fact that "In a triangle, if side length $a < b$, then $\angle A < \angle B$". (I don't know what the name of this is, anyone has an idea?)
2. This uniquely determines a Side Angle Angle triangle configuration in your parallelogram.
3. Use Sine rule and Cosine rule to continue as normal.
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3. Use Sine rule and Cosine rule to continue as normal.  The sine rule suffices. – barto Dec 31 '12 at 12:32
@barto Agreed. I didn't' want to give too much away as he was asking how to solve it on his own, as I was wanted to explain how that ambiguous "larger side is 55cm" would come into play. – Calvin Lin Dec 31 '12 at 18:37

Let $AB=55cm$ and $\angle BAC = 30^0$ and $\angle DAC=50$

Now, $\angle DAB = \angle DAC + \angle BAC = 50 + 30 = 80^0$

Hence, $\angle ABC = 180^0 - \angle DAB = 180^0 - 80^0 = 100^0$

Now, in $\triangle ABC,\angle CAB = 180^0 - \angle ABC -\angle BAC = 180^0 - 100^0 - 30^0 = 50^0$

Applying, sine rule in $\triangle ABC$, we get, $$\frac {AB}{\sin \angle ACB} = \frac {AC}{\sin \angle ABC}$$

$$\frac{AB}{\sin 50^0} = \frac {AC}{\sin 100^0}$$

Therefore, $$AC = AB\frac{\sin 100^0}{\sin 50^0} = 55\frac{2\sin 50^0\cos 50^0}{\sin 50^0} = 110\cos 50^0$$

Thus, length of diagonal = $110\cos 50^0$.

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