Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(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:

enter image description here

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.

share|improve this question
1  
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
1  
Very nice formulation of your question. Welcome to SE! –  David Mitra Dec 4 '11 at 22:00
1  
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

2 Answers 2

up vote 4 down vote accepted

For part A, try counting the area of the parallelogram in two different ways, as suggested by Jim Belk.

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.

share|improve this answer

EDIT: please ignore - I misread the diagram.

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.