Question

I took this today and just want to know how to do it next time I get a problem like this

Answer 1

In the given figure,

In triangle DEF,

$\tan 45^{\circ}=\frac{12}{EF}\Rightarrow EF=12$$\sin 45^{\circ}=\frac{12}{DF}\Rightarrow\frac{1}{\sqrt[]{2}}=\frac{12}{DF}=DF=12\sqrt[]{2}$

The length of DF is

$12\sqrt[]{2}$

Now in triangle DFG,

$\tan 45^{\circ}=\frac{DF}{FG}=\frac{12\sqrt[]{2}}{FG}$$FG=12\sqrt[]{2}$

The length of FG is

$12\sqrt[]{2}$

Now apply Pythagoras theorem to determine DG,

$DG^2=(12\sqrt[]{2})^2+(12\sqrt[]{2})^2$$DG^2=288+288=576$$DG=\sqrt[]{576}=24$

The length of DG is 24.