How to find triangle height if I know its area and angles

Question

For example, there is a triangle $ABC$ with angles $\alpha = 45^\circ, \beta = 120 ^\circ, \gamma = 15^\circ$. $\text{Area } S = 15$. How to find all three heights of the triangle?

Answer 1

Okay, let angle CAB = 45, angle ABC = 120 and angle BAC = 15.

Drop a line from C that is perpendicular to AB and call the point of intersection D. CBD is a 60-30-90 triangle so. Call BC x, then BD = x/2. And CD (which is the height of both triangle CBD and ABC) is $x\sqrt{3}/2$.

Meanwhile Triangle ACD is a 45-45-90 triangle so AD = CD = $x\sqrt{3}/2$ so AB = $(x\sqrt{3} -1)/2$

And by Pythagorean Theoreom AC = $\sqrt{AD^2 + CD^2} = \sqrt{3x^2} = x\sqrt 3$.

So the area of triangle $ABC = 1/2(AB)(CD) = 1/2(x\sqrt{3}-1)/2*x\sqrt{3}/2) = 1/8(3 -\sqrt{3})x^2 = 15$.

$x = BC = \sqrt{120/(3 - \sqrt{3})}$

Ugh.

AB = $(\sqrt{120/(3 - \sqrt{3})})(\sqrt{3} -1)/2$

And AC = $(\sqrt{120/(3 - \sqrt{3})})\sqrt3$.

Answer 2

HINT: $$\frac{a}{sin45}=\frac{b}{sin120}=\frac{c}{sin15}=2R$$ so

$a=\sqrt{2}R$, $b=\sqrt{3}R$ , $c=\frac{(\sqrt{3}-1)R}{\sqrt{2}}$

Now substitute $a,b,c$ in $$abc=4R \Delta$$ from which you will get $R$ and hence $a,b,c$.