You are photographing a race at a track meet, sitting 50 feet from the center of a straightaway. A runner traveling at 20 feet per second passes in front of you. a) write a trigonometric equation to find the angle thete you have to turn your camera to photograph the runner t seconds later. assume the runner is still on the straightaway and traveling at a constant speed. b) sole the trigonometric question to find the angle theta you have to turn the camera to photograph the runner 3 seconds later. c) if time t is twice as long, is the measure of the angle you have to turn the camera twice as big?

Question

You are photographing a race at a track meet, sitting 50 feet from the center of a straightaway. A runner traveling at 20 feet per second passes in front of you. a) write a trigonometric equation to find the angle thete you have to turn your camera to photograph the runner t seconds later. assume the runner is still on the straightaway and traveling at a constant speed. b) sole the trigonometric question to find the angle theta you have to turn the camera to photograph the runner 3 seconds later. c) if time t is twice as long, is the measure of the angle you have to turn the camera twice as big?

Answer 1

draw a diagram. It is clear that

(a) tanθ = 20t/50

(b) solve when t=3

Since tan2θ≠2tanθ, (c) is clearly false.

Answer 2

a) To find the angle theta (θ) you need to turn your camera to photograph the runner t seconds later, we can use the concept of trigonometry and the properties of right triangles.

Since the runner is traveling at a constant speed, the distance covered by the runner in t seconds is given by the formula: distance = speed × time. In this case, the speed is 20 feet per second.

Let's denote the distance covered by the runner in t seconds as x. Therefore, we have x = 20t.

To find the angle theta, we need to consider the right triangle formed between your camera position, the runner, and the straightaway. The adjacent side of this triangle is the distance from your camera position to the center of the straightaway, which is 50 feet. The opposite side is the distance covered by the runner, denoted as x.

Using the trigonometric definition of the tangent function (tan(theta) = opposite/adjacent), we can write the trigonometric equation as:
tan(theta) = x/50.

b) To solve the trigonometric equation and find the angle theta when t = 3 seconds, we substitute the value of x from the given information (x = 20t) and solve for theta.

Substituting the values, we have:
tan(theta) = (20 * 3)/50
tan(theta) = 60/50
tan(theta) = 6/5

To find the value of theta, you can use the inverse tangent (arctan) function on both sides of the equation. Therefore,
theta = arctan(6/5)

Using a calculator or an online tool that provides trigonometric functions, evaluate arctan(6/5) to find the value of theta.

c) No, if the time t is twice as long, the measure of the angle you have to turn the camera will not be twice as big. The angle theta is dependent on the distance covered by the runner, not the time. Therefore, as long as the runner is still on the straightaway and traveling at a constant speed, the angle required to photograph the runner will remain the same, regardless of how much time has passed.