Question

question 16 (mandatory) (1 point) saved
determine the value of x to the nearest cm.

Explanation:

Step1: Find the measure of angle in triangle

In triangle \(BCD\), using the angle - sum property of a triangle (\(180^{\circ}\)), the third angle \(\angle CBD=180^{\circ}-(55^{\circ} + 115^{\circ})=10^{\circ}\). Then, \(\angle EBA = 180^{\circ}-(90^{\circ}+10^{\circ}) = 80^{\circ}\). In triangle \(EAB\), \(\angle AEB=180^{\circ}-(70^{\circ}+80^{\circ}) = 30^{\circ}\).

Step2: Use the sine rule

In triangle \(EAB\) and triangle \(BCD\) (but we use the right - angled triangle relation). Wait, another approach: consider the sine rule in a more complex way. Wait, better to use the sine rule in triangle \(EAB\) and triangle \(BCD\) relation. Wait, actually, note that in triangle \(BCD\), using the sine rule \(\frac{BD}{\sin115^{\circ}}=\frac{52}{\sin10^{\circ}}\), so \(BD=\frac{52\sin115^{\circ}}{\sin10^{\circ}}\approx\frac{52\times0.9063}{0.1736}\approx271.7\). Then in triangle \(EAB\), \(\angle EBA = 80^{\circ}\), \(\angle A = 70^{\circ}\), \(\angle AEB=30^{\circ}\). Using the sine rule \(\frac{x}{\sin80^{\circ}}=\frac{BD\sin70^{\circ}}{\sin30^{\circ}}\). Wait, no, a simpler way:
In triangle \(EAB\) and triangle \(BCD\) (after calculating angles). Wait, actually, using the sine rule in triangle \(EAB\) and triangle \(BCD\) (after proper angle calculations).
Alternatively, note that in triangle \(BCD\):
\(\angle CBD = 180-(55 + 115)=10^{\circ}\), \(\sin\angle CBD=\sin10^{\circ}\), \(\sin\angle BCD=\sin115^{\circ}\), by sine rule \(\frac{BD}{\sin115^{\circ}}=\frac{CD}{\sin10^{\circ}}\), \(BD=\frac{52\sin115^{\circ}}{\sin10^{\circ}}\)
In triangle \(EAB\), \(\angle EBA = 180-(90 + 10)=80^{\circ}\), \(\angle A = 70^{\circ}\), \(\angle AEB=180-(70 + 80)=30^{\circ}\)
By sine rule \(\frac{x}{\sin80^{\circ}}=\frac{BD\sin70^{\circ}}{\sin30^{\circ}}\)
First, \(BD=\frac{52\sin115^{\circ}}{\sin10^{\circ}}\approx\frac{52\times0.9063}{0.1736}\approx271.7\)
Then \(\frac{x}{\sin80^{\circ}}=\frac{271.7\sin70^{\circ}}{\sin30^{\circ}}\), \(x=\frac{271.7\sin70^{\circ}\sin80^{\circ}}{\sin30^{\circ}}\approx\frac{271.7\times0.9397\times0.9848}{0.5}\approx 500\)

Answer:

\(x = 500\)