Question

1. a ladder rests against a wall. the top of the ladder reaches 12 feet up the wall, and makes an angle of 48° with the wall. how long is the ladder?
2. a ladder leans against a brick wall. the foot of the ladder is 8 feet from the wall. the ladder reaches a height of 25 feet on the wall. find the angle the ladder makes with the wall.
3. from the top of a fire tower, a forest ranger sees his partner on the ground at an angle of depression of 36°. if the tower is 40 feet in height, how far is the partner from the base of the tower?
4. an airplane is preparing to land at the airport. the airplane is flying at an altitude of about 2100 feet and is approximately 14,000 feet from the touchdown point on the runway. what angle does the approach path make with the runway?

Explanation:

Step1: Use trigonometric ratio

For problem 20:
We know that in a right - triangle (formed by the wall, ground and ladder), if the angle with the wall is \(\theta = 48^{\circ}\) and the adjacent side (height on the wall) \(a = 12\) feet, and we want to find the hypotenuse (length of the ladder) \(x\).
We use the cosine ratio \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), so \(\cos48^{\circ}=\frac{12}{x}\).
Then \(x=\frac{12}{\cos48^{\circ}}\).
Using a calculator, \(\cos48^{\circ}\approx0.6691\), so \(x=\frac{12}{0.6691}\approx17.9\) feet.

For problem 21:
In a right - triangle (formed by the wall, ground and ladder), the opposite side \(o = 8\) feet and the adjacent side \(a = 25\) feet.
We use the tangent ratio \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), so \(\tan\theta=\frac{8}{25}\).
Then \(\theta=\tan^{- 1}(\frac{8}{25})\).
Using a calculator, \(\theta\approx17.7^{\circ}\).

For problem 22:
In a right - triangle (formed by the tower, ground and the line of sight), the opposite side \(o = 40\) feet and we want to find the adjacent side \(x\).
Since the angle of depression is \(36^{\circ}\), the angle at the ground is also \(36^{\circ}\) (alternate interior angles).
We use the tangent ratio \(\tan36^{\circ}=\frac{40}{x}\), so \(x = \frac{40}{\tan36^{\circ}}\).
Using a calculator, \(\tan36^{\circ}\approx0.7265\), so \(x=\frac{40}{0.7265}\approx55.1\) feet.

For problem 23:
In a right - triangle (formed by the altitude, the horizontal distance to the touchdown point and the approach path), the opposite side \(o = 2100\) feet and the hypotenuse \(h = 14000\) feet.
We use the sine ratio \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), so \(\sin\theta=\frac{2100}{14000}=0.15\).
Then \(\theta=\sin^{-1}(0.15)\approx8.6^{\circ}\).

Answer:

1. Approximately \(17.9\) feet.
2. Approximately \(17.7^{\circ}\).
3. Approximately \(55.1\) feet.
4. Approximately \(8.6^{\circ}\).