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 triangle DEF

In right - triangle \(DEF\), \(\sin35^{\circ}=\frac{DF}{DE}\) (where \(DE\) is the hypotenuse, but we know \(EF = 40\) cm). Using the formula \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), we have \(\sin35^{\circ}=\frac{DF}{DE}\) and \(\cos35^{\circ}=\frac{EF}{DE}\). First, find \(DF\): \(DF = EF\tan35^{\circ}\) (since \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), in \(\triangle DEF\), \(\tan35^{\circ}=\frac{DF}{EF}\)). So \(DF = 40\times\tan35^{\circ}\approx40\times0.7002 = 28.008\) cm.

Step2: Find the length of \(x\) in triangle DFG

In right - triangle \(DFG\), \(\sin45^{\circ}=\frac{DF}{x}\) (where \(x\) is the hypotenuse of \(\triangle DFG\)). Since \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.707\), and \(DF\approx28.008\) cm. Then \(x=\frac{DF}{\sin45^{\circ}}\). Substitute \(DF = 28.008\) cm into the formula: \(x=\frac{28.008}{0.707}\approx39.6\) cm.

Answer:

A. \(39.6\) cm