Answer
Chapter 2 - Section 2.3 Problem Set: 56 (Answer)
Refer to Figure 1
The angle formed by diagonals DE and DG
is 35.3$^{\circ}$ (To the nearest tenths).
Work Step by Step
Chapter 2 - Section 2.3 Problem Set: 56 (Solution)
Refer to Figure 1
In $\triangle$DCG,
By Pythagoras' Theorem
$DG^2$ = $DC^2$ + $GC^2$
$DG^2$ = $3^2$ + $3^2$
$DG$ = $\sqrt{18}.$
In $\triangle$EDG,
$\tan \angle EDG$ = $\frac{EG}{DG}$
$\tan \angle EDG$ = $\frac{3}{\sqrt{18}}$
$\angle EDG$ = $\tan^{-1}(\frac{3}{\sqrt{18}})$
$\angle EDG$ = 35.3$^{\circ}$ (to the nearest tenths).
Therefore, the angle formed by diagonals DE and DG
is 35.3$^{\circ}$ (to the nearest tenths).
