An observer 150 feet from point C, where a weather balloon has been released. The balloon rises vertically at a rate of 8ft/sec/. At the very instant the balloon is 250 above ground how far is it from the observer? How Quickly is it moving away from the observer at that moment?

All answers I see for problems like this include sin tan cos but we did not learn this in class so I do not know how to solve this problem.

The way you worded the question,

"How quickly is it moving away from the observer at that moment" , implies to find the rate at which the distance between the balloon and the observer changes.
So we could just use Pythagoras

after t sec, height of balloon is 8t ft
so if d is the distance between them, (the hypotenuse)
d^2 = 150^2 + (8t)^2
2d dd/dt = 2(8t)(8) = 128t
dd/dt = 64/d

when 8t = 250
t = 250/8 = 125/4
d^2 = 22500 + 62500
d = √8500 = 10√85

dd/dt = 64/(10√85) = appr .694.. ft/sec

To solve this problem without using trigonometric functions, we can use the concept of similar triangles. Let's break down the problem step by step.

1. First, let's draw a diagram to visualize the situation. Place an observer at point C, and draw a vertical line representing the weather balloon's ascent.

| \
| \
| 250ft
| \
| \
| \
| \
| \
| \

2. From the given information, we know that the balloon is rising at a rate of 8 ft/sec. This means that for every second that passes, the balloon's height increases by 8 feet.

3. We are also given that the balloon is 150 feet away from the observer at point C. This forms a right triangle with the height and distance as its sides.

4. Now, let's focus on the moment when the balloon is 250 feet above the ground. We want to find how far it is horizontally from the observer at this point.

5. To find the horizontal distance, let's represent it as "x" (in feet).

Using the concept of similar triangles, we can set up the following proportion:

(distance from observer) / (height of balloon) = (distance from observer + x) / (height of balloon + 250)

Plugging in the known values:

150 / 0 = (150 + x) / 250

Simplifying the equation:

150 * 250 = (150 + x) * 0

37500 = 150 + x

x = 37500 - 150

x = 37350

Therefore, the balloon is approximately 37350 feet horizontally away from the observer at that moment.

6. Next, let's determine how quickly the balloon is moving away from the observer.

We know the balloon is rising at a rate of 8 ft/sec, and we found that it is approximately 37350 feet horizontally away. Thus, we can calculate the rate at which it is moving away:

Rate of horizontal movement = (Rate of vertical movement) * (Distance from observer)

Rate of horizontal movement = 8 ft/sec * 37350 ft

Rate of horizontal movement ≈ 298,800 ft/sec

Therefore, the balloon is moving away from the observer at a rate of approximately 298,800 feet per second at that moment.