Question

to find the distance between a point x and an inaccessible point z, a line segment xy is constructed. measurements show that xy = 962 m, angle xyz = 29°48, and angle yzx = 117°24. find the distance between x and z to the nearest meter. the distance between x and z is m. (do not round until the final answer. then round to the nearest meter as needed.)

Explanation:

Step1: Convert angles to decimal degrees

• Convert \(29^{\circ}48'\) to decimal degrees: \(48\div60 = 0.8\), so \(29^{\circ}48'=29.8^{\circ}\)
• Convert \(117^{\circ}24'\) to decimal degrees: \(24\div60 = 0.4\), so \(117^{\circ}24'=117.4^{\circ}\)
• Find angle \(X\) using the angle - sum property of a triangle (\(A + B + C=180^{\circ}\)). Let \(A = X\), \(B = Y=29.8^{\circ}\), \(C = Z = 117.4^{\circ}\). Then \(X=180-(29.8 + 117.4)=32.8^{\circ}\)

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). In \(\triangle XYZ\), if \(a = XZ\), \(b = XY = 962\) (opposite angle \(Z\)), \(A=\angle Y = 29.8^{\circ}\), \(B=\angle Z=117.4^{\circ}\)
We want to find \(XZ\). Using \(\frac{XZ}{\sin Y}=\frac{XY}{\sin Z}\)
Substitute the values: \(XZ=\frac{XY\times\sin Y}{\sin Z}\)
\(\sin Y=\sin(29.8^{\circ})\approx0.497\), \(\sin Z=\sin(117.4^{\circ})\approx0.887\), \(XY = 962\)
\(XZ=\frac{962\times0.497}{0.887}\)

Step3: Calculate the value

\(XZ=\frac{962\times0.497}{0.887}=\frac{478.114}{0.887}\approx539\)

Answer:

\(539\)