Question

a balloonist is directly above a straight road 1.8 miles long that joins two towns. she finds that the town closer to her is at an angle of depression of 40° and the farther town is at an angle of depression of 36°. how high above the ground is the balloon? the balloonist is about (square) 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 \( h \) be the height of the balloon above the ground. Let \( x \) be the distance from the closer town to the point on the road directly below the balloon. Then the distance from the farther town to that point is \( 1.8 - x \).
We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For the angle of depression of \(40^{\circ}\) (the angle of elevation from the closer town), \(\tan40^{\circ}=\frac{h}{x}\), so \(x = \frac{h}{\tan40^{\circ}}\). For the angle of depression of \(36^{\circ}\) (the angle of elevation from the farther town), \(\tan36^{\circ}=\frac{h}{1.8 - x}\), so \(1.8 - x=\frac{h}{\tan36^{\circ}}\).

Step2: Substitute \(x\)

Substitute \(x=\frac{h}{\tan40^{\circ}}\) into \(1.8 - x=\frac{h}{\tan36^{\circ}}\).
We get \(1.8-\frac{h}{\tan40^{\circ}}=\frac{h}{\tan36^{\circ}}\).
Since \(\tan40^{\circ}\approx0.8391\) and \(\tan36^{\circ}\approx0.7265\), the equation becomes \(1.8-\frac{h}{0.8391}=\frac{h}{0.7265}\).

Step3: Solve for \(h\)

Multiply through by \(0.8391\times0.7265\) to clear the fractions:
\(1.8\times0.8391\times0.7265-0.7265h = 0.8391h\).
\(1.8\times0.8391\times0.7265=(0.8391 + 0.7265)h\).
First, calculate \(1.8\times0.8391\times0.7265\approx1.8\times0.6093\approx1.0967\).
And \(0.8391+ 0.7265 = 1.5656\).
Then \(h=\frac{1.0967}{1.5656}\approx0.70\).

Answer:

\(0.70\)