Question

determine the value of x to the nearest tenth of a cm.
a) 39.6 cm
b) 40.2 cm
c) 38.7 cm
d) 37.1 cm

Explanation:

Step1: Find the length of \( DF \)

In right - triangle \( DEF \), \(\sin35^{\circ}=\frac{DF}{DE}\), \(EF = 40\) cm.
\(\sin35^{\circ}=\frac{DF}{40}\), so \(DF = 40\times\sin35^{\circ}\approx40\times0.5736 = 22.944\) cm.

Step2: Find the length of \( FG \)

Since \(EF + FG=EG\) and \(EF = 40\) cm, in right - triangle \( DFG \), \(\sin45^{\circ}=\frac{DF}{x}\) (where \(x = DG\)).
We know \(DF\approx22.944\) cm and \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7071\).
From \(\sin45^{\circ}=\frac{DF}{x}\), we can solve for \(x\): \(x=\frac{DF}{\sin45^{\circ}}\).
Substitute \(DF = 40\times\sin35^{\circ}\) into the formula: \(x=\frac{40\times\sin35^{\circ}}{\sin45^{\circ}}\).
\(x=\frac{40\times0.5736}{0.7071}\approx32.4\) (This is wrong. Let's use another way.

Since \( \triangle DEF\) and \( \triangle DFG\) are right - triangles and \(DE = DG\) (by symmetry, assume \(DE = DG\)).
First, in \(\triangle DEF\), \(\cos35^{\circ}=\frac{EF}{DE}\), \(EF = 40\) cm.
\(DE=\frac{40}{\cos35^{\circ}}\approx\frac{40}{0.8192}\approx48.83\) (wrong approach.

Correct way:
Since \( \triangle DEF\) and \( \triangle DFG\) are right - triangles and \(DF\) is common.
In \(\triangle DEF\), \(DF = EF\times\tan35^{\circ}\approx40\times0.7002 = 28.008\)
In \(\triangle DFG\), \(x=\frac{DF}{\sin45^{\circ}}\)
\(x=\frac{40\times\tan35^{\circ}}{\sin45^{\circ}}\)
\(\tan35^{\circ}\approx0.7002\), \(\sin45^{\circ}\approx0.7071\)
\(x=\frac{40\times0.7002}{0.7071}\approx39.6\)

Answer:

A. \(39.6\) cm