# Parallelogram trigonometry

(Sorry for the ambiguous title, couldn't think of a better one)

While leafing through a highschool textbook, I found what looked like an interesting question in trigonometry. My trigonometry skills are borderline 0, but I didn't expect it to be too much of a challenge. Well, I was wrong:

The sides of a parallelogram are $a$ and $b$ and its sharp angle is $\alpha$. The diagnols are $n$ and $m$, and the sharp angle between them is $\beta$.

A. Prove: $\frac{mn}{2ab} = \frac{\sin\alpha}{\sin\beta}$

B. Let: $\alpha = \beta$, $a < b$, $m < n$

Prove: $6a^2 + 2b^2 = 3m^2+n^2$

And in (rough) drawing: Following the law of cosines (and that $\cos(180-\theta) = -\cos(\theta)$):

$n^2 = a^2 + b^2 - 2ab \cos\alpha$ (in $\Delta ABC$)

$m^2 = a^2 + b^2 - 2ab \cos(180-\alpha) = a^2 + b^2 + 2ab \cos(\alpha)$ (in $\Delta DAC$)

$a^2 = (\frac{m}{2})^2 + (\frac{n}{2})^2 -2 \frac{m}{2} \frac{n}{2} \cos(\beta)$ (in $\Delta AEB$)

$b^2 = (\frac{m}{2})^2 + (\frac{n}{2})^2 -2 \frac{m}{2} \frac{n}{2} \cos(180 - \beta)$ (in $\Delta BEC$)

Expanding the last two equations:

$$a^2 = \frac{m^2}{4} + \frac{n^2}{4} - \frac{mn \cos(\beta)}{2}$$

$$b^2 = \frac{m^2}{4} + \frac{n^2}{4} + \frac{mn \cos(\beta)}{2}$$

$$\Rightarrow$$

$$a^2 + b^2 = \frac{m^2}{2} + \frac{n^2}{2}$$

And that's where I hit a wall. I have six variables, and can't find a way to express them in a fashion which resembles the end result. A major setback is that I couldn't find a way to express both alpha and beta in the same triangle - if I could, then the law of sines will probably be a rescuer.

If possible, I'd like that instead of solving it, maybe you can show me a guideline - where I went wrong, or what I'm missing. Thank you in advance.

• As a hint for question A, you could try applying the law of sines to two triangles with a common angle. You should end up with two equations for the sine of that common angle. – Heike Dec 4 '11 at 21:47
• Very nice formulation of your question. Welcome to SE! – David Mitra Dec 4 '11 at 22:00
• For part A, you can use the formula $\frac{1}{2} ab\sin\theta$ for the area of a triangle. – Jim Belk Dec 5 '11 at 0:17

For part B, notice that your diagram has $n$ and $m$ reversed, since $m<n$. In particular, $\alpha$ should be opposite $m$, not $n$. The modified version of your formula for $m$ is then $$m^2 = a^2+b^2 -2ab\cos\alpha.$$
Try combining this with your formula $$4a^2 = m^2 + n^2 -2mn\cos\beta$$ and use the result from part A.
The formula in A is wrong. Easiest way to see this may be to take the case where the figure is a square of side 1. Then $m=n=\sqrt2/2$, so the left side is $1/4$. $\alpha$ and $\beta$ are both right angles, so the right side is 1.
• In a unit square $m=n=\sqrt{2}$. – user7530 Dec 5 '11 at 11:58
• Ah, I misread the diagram - I thought $m$ was DE and $n$ was AE. Never mind. – Gerry Myerson Dec 5 '11 at 23:47