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

1. ## 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

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
Please supply ample small 'steps' in your answer. Lets me and others understand how you came to your answer.

2. Originally Posted by samtoucan24
Hi Guys

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

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
Please supply ample small 'steps' in your answer. Lets me and others understand how you came to your answer.
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, $\displaystyle h = b\tan{\theta}$

$\displaystyle A = \frac{bh}{2}$

$\displaystyle A = \frac{b^2\tan{\theta}}{2}$

$\displaystyle b = \sqrt{\frac{2A}{\tan{\theta}}}$