# Trig problem I'm stuck on

• Nov 1st 2009, 05:22 AM
Eleven Eleven
Trig problem I'm stuck on
I have a triangle that has internal angles of 35, 80 and 65 degrees. The area of the triangle needs to be 4000sq metres. Say I need to fence this triangle. Will 285 metres of fencing material be enough? If this is not enough fencing material can I change the side lengths of the triangle such that the area is still 4000sq m and no more than 285m of material is used? I'm not able to change the angles in this triangle by more than +or- 25 degrees.
Even just a suggestion as to where to start tackling this problem would be greatly appreciated.
• Nov 1st 2009, 08:22 AM
Hello Eleven Eleven
Quote:

Originally Posted by Eleven Eleven
I have a triangle that has internal angles of 35, 80 and 65 degrees. The area of the triangle needs to be 4000sq metres. Say I need to fence this triangle. Will 285 metres of fencing material be enough? If this is not enough fencing material can I change the side lengths of the triangle such that the area is still 4000sq m and no more than 285m of material is used? I'm not able to change the angles in this triangle by more than +or- 25 degrees.
Even just a suggestion as to where to start tackling this problem would be greatly appreciated.

I don't think you can quite do this with 285 m - the mimimum length of fencing you need will be 288 m.

Here's the calculation.

Suppose the triangle is $\displaystyle ABC$ with $\displaystyle A = 35, B = 80, C = 65$.

The area of the triangle is $\displaystyle \Delta= \tfrac12ab\sin C$, and using the Sine Rule $\displaystyle b = \frac{a\sin B}{\sin A}$. So
$\displaystyle \Delta = \frac12\frac{a^2\sin B \sin C}{\sin A}$

$\displaystyle \Rightarrow a^2 = \frac{2\Delta\sin A}{\sin B \sin C}$

$\displaystyle =\frac{8000 \sin 35}{\sin 80\sin65}$

$\displaystyle \Rightarrow a = 71.70$
This gives $\displaystyle b = \frac{a\sin B}{\sin A}=123.11$ and $\displaystyle c=\frac{a\sin C}{\sin A}=113.29$. The perimeter is then $\displaystyle 308.1$ m.

The minimum perimeter for a given area will be when the triangle is equilateral, all angles then being $\displaystyle 60^o$. But this gives each side as $\displaystyle 96.11$ m, with a perimeter of just over $\displaystyle 288$ m.