Question

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

Explanation:

Step1: Find $m\angle E$

In a right - triangle, the sum of angles is $180^{\circ}$. Given $\angle F = 90^{\circ}$ and $\angle G=45^{\circ}$, so $m\angle E=180^{\circ}-90^{\circ} - 45^{\circ}=45^{\circ}$

Step2: Use trigonometric ratios

Since $\sin G=\frac{EF}{EG}$ and $\cos G=\frac{FG}{EG}$, and $FG = 6\sqrt{6}\sin45^{\circ}$, $\sin45^{\circ}=\frac{\sqrt{2}}{2}$, $FG = 6\sqrt{3}$. Also, using the Pythagorean theorem in a $45 - 45-90$ triangle (but here we can use $\sin45^{\circ}=\frac{EF}{EG}$ and $\cos45^{\circ}=\frac{FG}{EG}$, since $FG = 6\sqrt{6}\cos45^{\circ}=6\sqrt{3}$ and from $\sin45^{\circ}=\frac{EF}{EG}$, cross - multiplying gives $EF = FG = 6\sqrt{3}$ (because in a $45 - 45-90$ triangle, the legs are equal when using the trigonometric ratios with the hypotenuse $EG$.

Another way:
In a right - triangle $\triangle EFG$ with $\angle F = 90^{\circ}$, $\angle G = 45^{\circ}$, $\angle E=45^{\circ}$ (so it is a $45 - 45-90$ triangle in terms of angle properties, but wait no, wait $\sin G=\frac{EF}{EG}$, $\cos G=\frac{FG}{EG}$. Given $FG = 6\sqrt{6}\cos45^{\circ}=6\sqrt{3}$, and since $\sin45^{\circ}=\frac{EF}{EG}$, and $EG$ is the hypotenuse. Using the Pythagorean theorem $EG=\sqrt{EF^{2}+FG^{2}}$, but since $\angle E=\angle G = 45^{\circ}$, $EF = FG = 6\sqrt{3}$ (by angle - side relationship in a triangle: equal angles have equal opposite sides), and $EG=\sqrt{(6\sqrt{3})^{2}+(6\sqrt{3})^{2}}=\sqrt{36\times3 + 36\times3}=\sqrt{72\times3}=\sqrt{216}=6\sqrt{6}$

Answer:

$m\angle E = 45^{\circ}$, $EF = 6\sqrt{3}$, $EG = 6\sqrt{6}$