# Solve for x using knowledge of triangles

#### skeeter

Math Team
Label the common vertical height of all three triangles as $h$ and the distance from the right angle to angle $x$ as $b$.

The large triangle is a 30-60-90 triangle. As such, $h\sqrt{3} = b+2 \implies h = \dfrac{b+2}{\sqrt{3}}$

For the medium triangle, $h=(b+1)\tan(40)$

Using these two equations, one can solve for $b$, and then determine a value for $h$.

Finally, using the small triangle, $\tan{x} = \dfrac{h}{b} \implies x = \arctan\left(\dfrac{b}{h}\right)$

#### skeeter

Math Team
typo correction ... $x=\arctan\left(\dfrac{h}{b}\right)$

#### skipjack

Forum Staff
As $h\tan(60^\circ) - h\tan(50^\circ) = 1,\ h = 1/(\tan(60^\circ) - \tan(50^\circ)) = 1.850833...$, etc.

One gets $$\displaystyle \tan(x) = \frac{1}{2\tan(50^\circ) - \tan(60^\circ)}$$.

#### mjsilverfly

typo correction ... $x=\arctan\left(\dfrac{h}{b}\right)$
Great, thank you! I understood everything you said. Following those steps, I got x = 56.9 degrees and that's the correct answer. YAY!