# Trig/Calculus - Given area and angle find sides of triangle

• Aug 29th 2009, 02:57 AM
samtoucan24
Trig/Calculus - Given area and angle find sides of triangle
Hi Guys

Its been 20 years since I looked at trig and calculus. Take it easy on me please (Happy)

In a right angled triangle, is it possible to find the size of the sides given
1 acute angle, and
the area of the triangle?

I think it is but I cant put the pieces together. I suspect the answer lies in calculus.

Imagine a triangle sloping up left to right. The 90 degree corner is the bottom right. The acute angle is on the bottom left.
The side I am interested in is the vertical side above the 90 degree corner. Its the Opposite side to the acute angle.

Any help much appreciated

Cheers

Sam

PS
• Aug 29th 2009, 03:18 AM
skeeter
Quote:

Originally Posted by samtoucan24
Hi Guys

Its been 20 years since I looked at trig and calculus. Take it easy on me please (Happy)

In a right angled triangle, is it possible to find the size of the sides given
1 acute angle, and
the area of the triangle?

I think it is but I cant put the pieces together. I suspect the answer lies in calculus.

Imagine a triangle sloping up left to right. The 90 degree corner is the bottom right. The acute angle is on the bottom left.
The side I am interested in is the vertical side above the 90 degree corner. Its the Opposite side to the acute angle.

Any help much appreciated

Cheers

Sam

PS
let $\displaystyle b$ be the base, $\displaystyle h$ the height, $\displaystyle \theta$ the given acute angle, and $\displaystyle A$ the given area.
$\displaystyle \tan{\theta} = \frac{h}{b}$
so, $\displaystyle h = b\tan{\theta}$
$\displaystyle A = \frac{bh}{2}$
$\displaystyle A = \frac{b^2\tan{\theta}}{2}$
$\displaystyle b = \sqrt{\frac{2A}{\tan{\theta}}}$