Question

solve the right triangle.
write your answers in simplified, rationalized form. do not round.
fg=
ef=
m∠g=°

Explanation:

Step1: Find $m\angle G$

In a right - triangle, the sum of angles is $180^{\circ}$. Since $\angle E = 90^{\circ}$ and $\angle F=60^{\circ}$, then $m\angle G=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$.

Step2: Use trigonometric ratios to find $FG$

We know that $\cos F=\frac{EF}{FG}$ and $\sin F=\frac{EG}{FG}$. Also, for a right - triangle with angles $30^{\circ}-60^{\circ}-90^{\circ}$, if the side opposite to $30^{\circ}$ is $x$, the side opposite to $60^{\circ}$ is $x\sqrt{3}$, and the hypotenuse is $2x$. Here, $EG = 4\sqrt{39}$. Let $EF=x$ and $FG = 2x$. Using the Pythagorean theorem $EG^{2}+EF^{2}=FG^{2}$. Substitute $EG = 4\sqrt{39}$, $EF=x$, and $FG = 2x$ into $a^{2}+b^{2}=c^{2}$ (where $a = EF$, $b = EG$, $c = FG$).

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But using the sine ratio: $\sin F=\frac{EG}{FG}$. Since $\angle F = 60^{\circ}$ and $EG = 4\sqrt{39}$, and $\sin60^{\circ}=\frac{\sqrt{3}}{2}$. Let $FG = y$. Then $\sin60^{\circ}=\frac{4\sqrt{39}}{y}$, so $y=\frac{4\sqrt{39}}{\sin60^{\circ}}=\frac{4\sqrt{39}}{\frac{\sqrt{3}}{2}}=\frac{8\sqrt{39}}{\sqrt{3}}=8\sqrt{13}$.

Step3: Find $EF$

Using the cosine ratio: $\cos F=\frac{EF}{FG}$. Since $\angle F = 60^{\circ}$ and $FG = 8\sqrt{13}$, and $\cos60^{\circ}=\frac{1}{2}$. Let $EF = z$. Then $z = FG\times\cos60^{\circ}=8\sqrt{13}\times\frac{1}{2}=2\sqrt{13}$.

Answer:

$FG = 8\sqrt{13}$; $EF = 2\sqrt{13}$; $m\angle G = 30^{\circ}$