Question

find the missing length(s).
3.
4.
9.2 special right triangles

Explanation:

Step1: Solve for $x$ (30-60-90 triangle)

Use $\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}$
$\tan(60^\circ)=\frac{\sqrt{3}}{x} \implies x=\frac{\sqrt{3}}{\tan(60^\circ)}$
Since $\tan(60^\circ)=\sqrt{3}$, $x=\frac{\sqrt{3}}{\sqrt{3}}=1$

Step2: Solve for $y$ (30-60-90 triangle)

Use $\sin(60^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}$
$\sin(60^\circ)=\frac{\sqrt{3}}{y} \implies y=\frac{\sqrt{3}}{\sin(60^\circ)}$
Since $\sin(60^\circ)=\frac{\sqrt{3}}{2}$, $y=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=2$

Step3: Solve for $h$ (right triangle segment)

Use Pythagorean theorem: $h^2 + 2^2 = 4^2$
$h^2=16-4=12 \implies h=\sqrt{12}=2\sqrt{3}$

Answer:

For problem 3: $x=1$, $y=2$
For problem 4: $h=2\sqrt{3}$