How can I solve a triangle knowing its area, one side and an opposite angle?

Question

I need to calculate the missing elements of a triangle, knowing its area one side and the angle opposite the given side. The triangle is not a right angle triangle, nor is it equilateral or isosceles.

Can someone advise a formula?

Answer 1

Treat the side as the chord of a circle in which it subtends the given angle - the extended sine rule gives the radius. Then use the area of the triangle to calculate the height and you can identify the correct point on the circle.

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Let the side you are given be $a$, and take this as the base of the triangle, and the angle opposite be $A$.

The extended version of the sine rule tells us that $\cfrac a{\sin A}=2R$ where $R$ is the circumradius of the triangle, so you can find $R$.

Having found $R$ you take the perpendicular bisector of your side, and locate one of the two points (above/below) which is distance $R$ from the extremities of the side (vertices $B, C$) and construct the circle.

With side $a$ as the base you calculate the height from area $=\cfrac {ah}2$.

Then take a line $l$ parallel to side $a$ and distance $h$ from it (on the correct side for the angle you want - if you get the wrong side the angle will be $180^{\circ}-A$). If the triangle is possible this will cut the circle in two points (you will see a symmetry about the perpendicular bisector) unless $l$ happens to be a tangent, when you get a single point. One of these points can be taken as vertex $A$.

I have not advised a formula, but a method. Following the method and doing the algebra will give you a formula if you need one.

Answer 2

The figure above shows the situation we have. Side $BC$ has a known length $s$, and $\angle A = \phi $ is known as well. Since the area $A$ is known, we can calculate the height $h$ as

$ h = \dfrac{2 A}{s} $

Next segment $BC$ into two segments $BF$ of length $x$ and $FC$ of length $s - x$,

Then

$ \phi = \tan^{-1} \dfrac{ x }{h} + \tan^{-1} \dfrac{s - x}{h} $

Apply the $\tan$ function to both sides

$ \tan \phi = \dfrac{ \dfrac{x}{h} + \dfrac{s - x}{h} }{ 1 - \dfrac{ x(s -x) }{h^2} } $

Simplifying,

$ \tan \phi ( x^2 - s x + h^2 ) = s h $

And this is a quadratic equation in $x$ which can be solved using the quadratic formula.

Once $x$ is known all the sides and the triangle angles become immediately known,

because,

$\angle B = \tan^{-1} \dfrac{h}{x} $

$\angle C = \tan^{-1} \dfrac{h}{s - x} $

$b = \sqrt{ h^2 + (s - x)^2 }$

$c = \sqrt{ h^2 + x^2 } $