Question

solve the right triangle. write your answers as integers or as decimals rounded to the nearest tenth. fg = eg = m∠f = °

Explanation:

Step1: Find $m\angle F$

In right - triangle $EFG$, $\cos\angle F=\frac{FG}{EF}$. Let $EF = 7$, assume $EG = FG = x$ (isosceles right - triangle property, we can also use trigonometric ratios). Using $\cos\angle F=\frac{FG}{EF}$, if we use the inverse cosine function. Let's use the Pythagorean theorem $EG^{2}+FG^{2}=EF^{2}$, since $EG = FG$, we have $2x^{2}=49$, $x^{2}=\frac{49}{2}$, $x=\frac{7}{\sqrt{2}}\approx4.95$. Also, $\sin\angle F=\frac{EG}{EF}$, $\sin\angle F=\frac{x}{7}$. Substituting $x = \frac{7}{\sqrt{2}}$, $\sin\angle F=\frac{1}{\sqrt{2}}\approx0.707$. Using $\angle F=\sin^{- 1}(0.707)\approx45.6^{\circ}$ (using a calculator, $\sin^{-1}(0.707)\approx45.6^{\circ}$)

Step2: Find $EG$ and $FG$

By Pythagorean theorem $EG^{2}+FG^{2}=EF^{2}$, and since $\angle G = 90^{\circ}$ and from $\angle F\approx45.6^{\circ}$, $\angle E=180^{\circ}-\angle F - \angle G\approx44.4^{\circ}\approx45^{\circ}$ (approximate, because of rounding), so the triangle is approximately isosceles. $EG^{2}+FG^{2}=49$, and $EG = FG$. Then $2EG^{2}=49$, $EG^{2}=24.5$, $EG=\sqrt{24.5}\approx4.95\approx5.0$, $FG=\sqrt{24.5}\approx4.95\approx5.0$

Answer:

$m\angle F = 45.6^{\circ}$, $EG = 5.0$, $FG = 5.0$