Trigonometry: Find the value of $x$

Question

I have been given this triangle and I am asked to find the value of $x$.

I'm not able to understand few things-

1) Is it a valid triangle? (Three angles could be $180^{\circ}$.)

2) How would I find $x$?

Answer 1

(1) Yes this is a "valid triangle". $$ \angle A = \angle B = 80^\circ, \quad \angle C=20^\circ. $$

(2) Choose the scale: let $AB=1$. Let point $D\in AC$, $\angle ABD=60^\circ$. Let point $E\in BC$, $\angle BAE=70^\circ$. So $x = \angle AED$.

We can find $AD$ and $AE$, by applying the law of sines to triangles $ABD$ and $ABE$: $$ {AD\over\sin60^\circ} = {AB\over\sin40^\circ}, \quad AD={AB\cdot\sin60^\circ\over\sin40^\circ} \approx{0.8660253\over0.6427876} \approx 1.347296, $$ $$ {AE\over\sin80^\circ} = {AB\over\sin30^\circ}, \quad AE={AB\cdot\sin80^\circ\over\sin30^\circ} \approx{0.98480775\over0.5} \approx 1.9696155. $$ Now we can find $DE$ using coordinates. Assume $A$ is the origin, $x_A=y_A=0$, then $$ x_D=AD\cos80^\circ\approx0.233955, \quad y_D=AD\sin80^\circ\approx1.3268275, $$ $$ x_E=AE\cos70^\circ\approx0.673648, \quad y_E=AE\sin70^\circ\approx1.8508331, $$ $$ DE=\sqrt{(x_E-x_D)^2+(y_E-y_D)^2}\approx\sqrt{0.439693^2+0.524006^2} =0.684041\ldots $$ Finally, we find $x$ by applying the law of sines to triangle $ADE$: $$ {AD\over\sin x} = {DE\over\sin 10^\circ}, \quad \sin x = {AD\cdot\sin10^\circ\over DE} \approx 0.342020, $$ $$ x = \arcsin(0.342020) \approx 20^\circ. $$