Question

1. kurt spots a bird sitting at the top of a 40 - foot - tall telephone pole. if the angle of elevation from the ground where he is standing to the bird is ( 59^{circ} ), how far is kurt standing from the base of the pole?
2. a helicopter flying 1,600 feet above the ground spots an airplane flying above. if the horizontal distance between the helicopter and airplane is 3,055 feet and the angle of elevation is ( 71^{circ} ), find the airplanes altitude.
3. a shark is swimming 28 feet below sea level. if the angle of depression from a boat on the water to the shark is ( 19^{circ} ), what is the horizontal distance between the boat and the shark?
4. while standing on top of a hill at a ski resort, lisa spots the ski lodge below. if the vertical height of the hill is 1209 feet and the angle of depression is ( 53^{circ} ), what distance must lisa ski in order to reach the lodge?

Explanation:

Step1: Use trigonometric ratio

For problem 13, we have a right - triangle where the opposite side to the angle of elevation (\(\theta = 59^{\circ}\)) is \(y = 40\) (height of the pole) and we need to find the adjacent side \(x\) (distance from Kurt to the base of the pole). We use the tangent ratio \(\tan\theta=\frac{y}{x}\).
So, \(\tan59^{\circ}=\frac{40}{x}\).

Step2: Solve for \(x\)

Since \(\tan59^{\circ}\approx1.6643\), then \(x=\frac{40}{\tan59^{\circ}}\).
\(x=\frac{40}{1.6643}\approx24\).

For problem 14, the horizontal distance \(x = 3055\) and the angle of elevation \(\theta=71^{\circ}\). The vertical distance (difference in altitude) \(y\) (airplane's altitude above the helicopter) is given by \(y=x\tan\theta\).
\(y = 3055\times\tan71^{\circ}\). Since \(\tan71^{\circ}\approx2.9042\), \(y=3055\times2.9042\approx8872.9\). The airplane's altitude is \(1600 + 8872.9=10472.9\).

For problem 15, the opposite side \(y = 28\) (depth of the shark) and the angle of depression (which is equal to the angle of elevation from the shark to the boat) \(\theta = 19^{\circ}\). Using \(\tan\theta=\frac{y}{x}\), we have \(\tan19^{\circ}=\frac{28}{x}\). Since \(\tan19^{\circ}\approx0.3443\), \(x=\frac{28}{0.3443}\approx81.3\).

For problem 16, the vertical height \(y = 1209\) and the angle of depression \(\theta = 53^{\circ}\). The angle of depression is equal to the angle of elevation from the lodge to Lisa. We use the sine ratio \(\sin\theta=\frac{y}{x}\) (where \(x\) is the distance Lisa needs to ski). So, \(\sin53^{\circ}=\frac{1209}{x}\). Since \(\sin53^{\circ}\approx0.7986\), \(x=\frac{1209}{0.7986}\approx1514\).

Answer:

1. \(24\)
2. \(10472.9\)
3. \(81.3\)
4. \(1514\)