Problem of congruence of triangles

Question

Find x.

Problem

Solutión 1

$2w+90+x=180\Rightarrow2w+x=90$

$w=\dfrac{90-x}{2}$

$\sqrt{a^{2}+3}=c$

$sen(w)=\dfrac{\sqrt{3}}{c}$

$sen(\dfrac{90-x}{2})=\dfrac{\sqrt{3}}{\sqrt{a^{2}+3}}$

$\dfrac{h}{2}=sen(x)$ $and$ $h=2sen(x)$

$\dfrac{h}{c}=sen(\dfrac{90-x}{2})$ $and$ $\sqrt{a^{2}+3}=c$

$\dfrac{h}{\sqrt{a^{2}+3}}=sen(\dfrac{90-x}{2})$

$\dfrac{2sen(x)}{\sqrt{a^{2}+3}}=sen(\dfrac{90-x}{2})$

$sen(\dfrac{90-x}{2})=\dfrac{\sqrt{3}}{\sqrt{a^{2}+3}}$

$2sen(x)=\sqrt{3}$

$sen(x)=\dfrac{\sqrt{3}}{2}$

$x=60°$

Solutión 2

$\begin{align}&\frac{\sqrt{3}}{sen(w)}=\frac{common side}{sen(90°)} \to > common side=\frac{\sqrt{3}}{sen(w)}\\&Además...\\&\frac{2}{sen(w)}=\frac{common side}{sen(x)}\\&\text{using the above...}\\&\frac{2}{sen(w)}=\frac{\sqrt{3}}{sen(w) \cdot > sen(x)} \to\\&sen(x)=\frac{\sqrt{3}}{2}\\&x = > arcsen(\frac{\sqrt{3}}{2})\\&x=60°\\&\end{align}$

How to find x ?, the subject is congruence of geometry triangles, but I have only been able to solve it using trigonometry, sines, and I have not been able to solve it using congruences, or similarities. Someone knows, I would appreciate your help.

Answer 1

At some point I think you need a little trig.

Lets call the length from $2w$ to the right angle $b$ and the hypotenuse $c$

One of our angle bisector theorems says $\frac {\sqrt{3}}{b} = \frac {2}{c}$

therefore $\frac{b}{c} = \frac {\sqrt {3}}{2} = \sin x$

Answer 2

Outline:

• Drop a perpendicular $DE$ from $D$ to side $AB$.$\\[4pt]$
• Argue that triangles $ACD$ and $AED$ are congruent.$\\[4pt]$
• Hence $DE=\sqrt{3}$.$\\[4pt]$
• Apply the Pythagorean theorem in triangle $DEB$ to get $BE=1$.$\\[4pt]$
• Deduce that triangle $DEB$ is a $30$-$90$-$60$ right triangle.