Question

15 points, solve a right triangle. Please simplify your answers as it says in the picture.

Answer 1

Given:

In triangle EFG,
`EF=10√(2), m\angle G=90^\circ,m\angle E=60^\circ`.

To find:

The measures of FG, EG and angle F.

Solution:

In triangle EFG,

`m\angle E+m\angle F+m\angle G=180^\circ` [Angle sum property]

`60^\circ+m\angle F+90^\circ=180^\circ`

`m\angle F+150^\circ=180^\circ`

`m\angle F=180^\circ-150^\circ`

`m\angle F=30^\circ`

In a right angle triangle,

`\sin \theta=(Perpendicular)/(Hypotenuse)`

`\sin E=(FG)/(EF)`

`\sin (60^\circ)=(FG)/(10√(2))`

`(√(3))/(2)* 10√(2)=FG`

`5√(6)=FG`

Similarly,

`\sin F=(EG)/(EF)`

`\sin (30^\circ)=(EG)/(10√(2))`

`(1)/(2)* 10√(2)=EG`

`5√(2)=EG`

Therefore, the required measures are
`FG=5√(6), EG=5√(2), m\angle F=30^\circ`.