Question

an aircraft (at z) is spotted by two observers (at x and y) who are l = 1750 feet apart. as the airplane passes over the line joining them, each observer takes a sighting of the angle of elevation to the plane, as indicated in the figure. if a = 35°, and b = 20°, how high is the airplane?
the elevation of the plane is approximately □ feet.
(round the final answer to two decimal places as needed. round all intermediate values to four decimal places as needed.)

Explanation:

Step1: Express \(XD\) and \(YD\) in terms of \(h\)

Using the tangent function in right - triangles \(XZD\) and \(YZD\).
In \(\triangle XZD\), \(\tan A=\frac{h}{XD}\), so \(XD = \frac{h}{\tan A}\).
In \(\triangle YZD\), \(\tan B=\frac{h}{YD}\), so \(YD=\frac{h}{\tan B}\).
Since \(L = XD + YD\), we have \(L=\frac{h}{\tan A}+\frac{h}{\tan B}\).
Factor out \(h\): \(L = h(\frac{1}{\tan A}+\frac{1}{\tan B})\).

Step2: Solve for \(h\)

We know that \(L = 1750\), \(A = 35^{\circ}\), and \(B=20^{\circ}\).
First, calculate \(\tan35^{\circ}\approx0.7002\) and \(\tan20^{\circ}\approx0.3640\).
Then \(\frac{1}{\tan A}+\frac{1}{\tan B}=\frac{1}{0.7002}+\frac{1}{0.3640}\approx1.4282 + 2.7473=4.1755\).
From \(L = h(\frac{1}{\tan A}+\frac{1}{\tan B})\), we can solve for \(h\) as \(h=\frac{L}{\frac{1}{\tan A}+\frac{1}{\tan B}}\).
Substitute \(L = 1750\) and \(\frac{1}{\tan A}+\frac{1}{\tan B}\approx4.1755\) into the formula: \(h=\frac{1750}{4.1755}\approx419.11\).

Answer:

\(419.11\)