Answer
Chapter 2 - Section 2.3 Problem Set: 57 (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: 57 (Solution)
Refer to Figure 1
In $\triangle$CDH,
By Pythagoras' Theorem
$CH^2$ = $CD^2$ + $DH^2$
$CH^2$ = $x^2$ + $x^2$
$CH$ = $x\sqrt{2}$
In $\triangle$FCH,
$\tan \angle FCH$ = $\frac{FH}{CH}$
$\tan \angle FCH$ = $\frac{x}{x\sqrt{2}}$
$\angle FCH$ = $\tan^{-1}(\frac{1}{\sqrt{2}})$
$\angle FCH$ = 35.3$^{\circ}$ (To the nearest tenths)
Therefore, the angle formed by diagonals CF and CH
is 35.3$^{\circ}$ (To the nearest tenths)
