1. **State the problem:** We are given a right triangle GHF with right angle at G. Given sides: - GH = $4\sqrt{17}$ - HF = $4\sqrt{34}$ We need to find: - FG (the third side) - $m\angle F$ - $m\angle H$ 2. **Use the Pythagorean theorem:** For a right triangle with legs $a$, $b$ and hypotenuse $c$, the relation is: $$c^2 = a^2 + b^2$$ Here, HF is the hypotenuse, so: $$HF^2 = GH^2 + FG^2$$ 3. **Calculate FG:** $$\begin{aligned} (4\sqrt{34})^2 &= (4\sqrt{17})^2 + FG^2 \\ 16 \times 34 &= 16 \times 17 + FG^2 \\ 544 &= 272 + FG^2 \\ FG^2 &= 544 - 272 = 272 \\ FG &= \sqrt{272} = \sqrt{16 \times 17} = 4\sqrt{17} \end{aligned}$$ 4. **Find angles $m\angle F$ and $m\angle H$ using trigonometry:** - $m\angle F$ is opposite side GH and adjacent side FG: $$\tan(m\angle F) = \frac{GH}{FG} = \frac{4\sqrt{17}}{4\sqrt{17}} = 1$$ $$m\angle F = \tan^{-1}(1) = 45^\circ$$ - Since the triangle is right angled and angles sum to 180°: $$m\angle H = 90^\circ - m\angle F = 90^\circ - 45^\circ = 45^\circ$$ **Final answers:** - $FG = 4\sqrt{17}$ - $m\angle F = 45^\circ$ - $m\angle H = 45^\circ$