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: Use trigonometric relations

Let \(x\) be the distance from \(X\) to \(D\) and \(y\) be the distance from \(Y\) to \(D\). We know that \(L=x + y=1750\). Also, \(\tan A=\frac{h}{x}\) (so \(x = \frac{h}{\tan A}\)) and \(\tan B=\frac{h}{y}\) (so \(y=\frac{h}{\tan B}\)).

Step2: Substitute \(x\) and \(y\) into \(L=x + y\)

Substituting \(x\) and \(y\) into \(L=x + y\), we get \(L=\frac{h}{\tan A}+\frac{h}{\tan B}=h(\frac{1}{\tan A}+\frac{1}{\tan B})\).
Given \(A = 35^{\circ}\), \(\tan A=\tan35^{\circ}\approx0.7002\); \(B = 20^{\circ}\), \(\tan B=\tan20^{\circ}\approx0.3640\).
Then \(L = h(\frac{1}{0.7002}+\frac{1}{0.3640})\).

Step3: Solve for \(h\)

Since \(L = 1750\), we have \(h=\frac{1750}{\frac{1}{0.7002}+\frac{1}{0.3640}}\).
First, calculate \(\frac{1}{0.7002}+\frac{1}{0.3640}\approx1.428 + 2.747=4.175\).
Then \(h=\frac{1750}{4.175}\approx419.16\).

Answer:

\(419.16\)