Given:
In triangle EFG, [tex]EF=10\sqrt{2}, m\angle G=90^\circ,m\angle E=60^\circ[/tex].
To find:
The measures of FG, EG and angle F.
Solution:
In triangle EFG,
[tex]m\angle E+m\angle F+m\angle G=180^\circ[/tex] [Angle sum property]
[tex]60^\circ+m\angle F+90^\circ=180^\circ[/tex]
[tex]m\angle F+150^\circ=180^\circ[/tex]
[tex]m\angle F=180^\circ-150^\circ[/tex]
[tex]m\angle F=30^\circ[/tex]
In a right angle triangle,
[tex]\sin \theta=\dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex]\sin E=\dfrac{FG}{EF}[/tex]
[tex]\sin (60^\circ)=\dfrac{FG}{10\sqrt{2}}[/tex]
[tex]\dfrac{\sqrt{3}}{2}\times 10\sqrt{2}=FG[/tex]
[tex]5\sqrt{6}=FG[/tex]
Similarly,
[tex]\sin F=\dfrac{EG}{EF}[/tex]
[tex]\sin (30^\circ)=\dfrac{EG}{10\sqrt{2}}[/tex]
[tex]\dfrac{1}{2}\times 10\sqrt{2}=EG[/tex]
[tex]5\sqrt{2}=EG[/tex]
Therefore, the required measures are [tex]FG=5\sqrt{6}, EG=5\sqrt{2}, m\angle F=30^\circ[/tex].
