## Trigonometry 7th Edition

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).
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).