Question

a balloonist is directly above a straight road 1.3 miles long that joins two towns. she finds that the town closer to her is at an angle of depression of 39° and the farther town is at an angle of depression of 38°. how high above the ground is the balloon? the balloonist is about \boxed{} miles above the ground. (round the final answer to two decimal places as needed. round all intermediate values to four decimal places as needed.)

Explanation:

Step1: Set up variables

Let the height of the balloon be \( h \) miles. Let the distance from the closer town to the point on the road directly below the balloon be \( x \) miles. Then the distance from the farther town to that point is \( 1.3 - x \) miles.
Since the angle of depression is equal to the angle of elevation from the town to the balloon. For the closer town, \(\tan39^{\circ}=\frac{h}{x}\), so \(x = \frac{h}{\tan39^{\circ}}\). For the farther town, \(\tan38^{\circ}=\frac{h}{1.3 - x}\), so \(1.3 - x=\frac{h}{\tan38^{\circ}}\).

Step2: Substitute \(x\)

Substitute \(x = \frac{h}{\tan39^{\circ}}\) into \(1.3 - x=\frac{h}{\tan38^{\circ}}\).
We get \(1.3-\frac{h}{\tan39^{\circ}}=\frac{h}{\tan38^{\circ}}\).

Step3: Solve for \(h\)

First, find the values of \(\tan38^{\circ}\approx0.7813\) and \(\tan39^{\circ}\approx0.8007\).
The equation becomes \(1.3-\frac{h}{0.8007}=\frac{h}{0.7813}\).
Multiply through by \(0.8007\times0.7813\) to clear the fractions:
\(1.3\times0.8007\times0.7813-0.7813h = 0.8007h\).
\(1.3\times0.8007\times0.7813=(0.8007 + 0.7813)h\).
\(1.3\times0.6251=(1.582)h\).
\(0.8126 = 1.582h\).
\(h=\frac{0.8126}{1.582}\approx0.51\).

Answer:

\(0.51\)