Answer
Chapter 2 - Section 2.3 Problem Set: 55 (Answer)
Refer to Figure 1
The angle formed by diagonals CF and CH
is 35.3$^{\circ}$ (To the nearest tenths)
Work Step by Step
Chapter 2 - Section 2.3 Problem Set: 55 (Solution)
Refer to Figure 1
In $\triangle$CDH,
by Pythagoras' Theorem,
$CH^2$ = $CD^2$ + $DH^2$
$CH^2$ = $5^2$ + $5^2$
$CH$ = $\sqrt{50}$
In $\triangle$FCH,
$\tan \angle FCH$ = $\frac{FH}{CH}$
$\tan \angle FCH$ = $\frac{5}{\sqrt{50} }$
$\angle FCH$ = $\tan^{-1}(\frac{5}{\sqrt{50}})$
$\angle FCH$ = 35.3$^{\circ}$ (To the nearest tenths)
Therefore, the angle formed by diagonals CF and CH
is 35.3$^{\circ}$ (To the nearest tenths)
