Question

solve the right triangle.

write your answers in simplified, rationalized form. do not round.
fh =

m∠h = °

m∠g = °

Explanation:

Step1: Calculate side $FH$ via Pythagoras

In right $\triangle FHG$, hypotenuse $HG=2\sqrt{94}$, leg $FG=2\sqrt{47}$. Use $FH=\sqrt{HG^2-FG^2}$.

$$\begin{align*} FH&=\sqrt{(2\sqrt{94})^2-(2\sqrt{47})^2}\\ &=\sqrt{4\times94 - 4\times47}\\ &=\sqrt{376 - 188}\\ &=\sqrt{188}\\ &=\sqrt{4\times47}\\ &=2\sqrt{47} \end{align*}$$

Step2: Find $m\angle G$ using cosine

$\cos G=\frac{FG}{HG}$, substitute values:

$$\begin{align*} \cos G&=\frac{2\sqrt{47}}{2\sqrt{94}}\\ &=\frac{\sqrt{47}}{\sqrt{2\times47}}\\ &=\frac{1}{\sqrt{2}}\\ &=\frac{\sqrt{2}}{2} \end{align*}$$

Thus $m\angle G=45^\circ$.

Step3: Calculate $m\angle H$

In right triangle, acute angles sum to $90^\circ$.
$$m\angle H=90^\circ - m\angle G=90^\circ-45^\circ=45^\circ$$

Answer:

$FH = 2\sqrt{47}$
$m\angle H = 45^\circ$
$m\angle G = 45^\circ$