Trig problem I'm stuck on

Eleven Eleven · #1 $Old$ November 1st, 2009, 07:22 AM

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.

Grandad · #2 $Old$ November 1st, 2009, 10:22 AM

Hello Eleven Eleven

Quote:

Originally Posted by Eleven Eleven $View Post$

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 $ABC$ with $A = 35, B = 80, C = 65$ .

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

$\Delta = \frac12\frac{a^2\sin B \sin C}{\sin A}$

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

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

$\Rightarrow a = 71.70$

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

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

Grandad

Eleven Eleven · #3 $Old$ November 1st, 2009, 04:37 PM

That's great. It's got me started so thank you!