Pre Calculus Question! Please help!?

Question

From a helicopter H feet above the ocean, the angle of depression to the pilot's horizon is A. Prove that,

cotA= R/ √(2RH+H²)

where, R is the radius of the Earth.


How do you solve this problem? I don't even know how to draw the picture! Please help!

Answer 1

Think of the planet, and a helicopter flying over it. Draw a line from the center of the earth up to the helicopter and think of this as a hypotenuse of length R+H of a right triangle we will draw. Call the center of the earth L, the helicopter will be point M.

The horizon is the point at which you can no longer see further along the earth because it's curved away, and the line to the horizon from the helicopter is a tangent line to the earth that intersects the helicopter. This is one of our legs of our right triangle. Call the point that touches the earth N.

Where our last line was tangent to the earth, draw a perpendicular line (giving us a 90 degree angle) to the center of the earth of length R.

We now have a right triangle, with angles L, M, and N, and lines opposite of l, m, and n. We know the lengths of m and n; length of m = R, length of n = R + H.

The angle of depression is the complement to our M angle. That is, angle M + the angle of depression equals 90. The cot(90 degrees - x) = tan(x), so to find the cot of our A, we need only find the tan of out M.

Tan is opposite over adjacent, which in our terms of l,m,n is m/l

We know m = R, but l we do net you know. From the pythagorean theorem we know that n^2 = l^2 + m^2

(H+R)^2 = l^2 + R^2
H^2 + 2RH + R^2 = l^2 + R^2
H^2 + 2RH = I^2
√(H^2 + 2RH) = I

Therefore tan(M) = R/√(H^2 + 2RH) = cot(A)

--charlie

Answer 2

This is difficult to visualize and that makes the challenge more of a drawing problem than a math problem. Here is a way to get started ...
1. Draw an arc representing the ocean, then place a dot representing the helicopter above that. The height of the helicoptor is (R+H) from the center of the earth (dumb, I know).

2. Draw a line straight out from the helicoptor (pilots line of sight when he looks straight ahead.) Make it an inche or so long. Now from the end of this line, draw a vertical line down to the ocean (right angle).

3. Now draw a line from the helicopter to where the vertical line meets the ocean. This will be the hypotenuse of your triangle and the angle where the helicopter is will be A.

4. Now you have a triangle to work with. The vertical line is now H long and angle A is defined.

5. There is also another triangle from the helicopter to the center of the earth and back to the point the pilot's line of vision hits the ocean.

6. Use the two triangles to get started with tangent and cotangent definitions ...

Unfortunately, unless this helicoptor is really high, the earth's radius is so big, H hardly matters -- unless, of course, you are in the helicopter, then H is the only thing that matters.

Answer 3

Drawing the question is the most difficult part, and I have to admit to be stucking as well.

I imagine that there's a line drawn from the helicopter down to the horizon, making a right triangle with a line drawn to the helicopter from the center of the earth, but then the answer is cot(A) = R+H/R.

Answer 4

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Answer 5

Draw a circle with radius = R from O the center of the circle. Extend the radius a distance H to the point C. Now draw a tangent to the circle from C to D. Draw another radius to the point D. Draw a line perpendicular to OC and let it intersect the radius OD extended at E. There is your picture. The angle ECD is the angle of depression (A) and is also equal
to the angle COE.

If you have drawn the picture correctly, you should see that cot A = (R + H)/CE. So you need to use your brain to express CE in terms of R and H to get your answer. I have to leave you with something to do yourself.

Answer 6

Here's your picture (maximize it for the best resolution):

http://bestshotammo.com/str/helicopter.jpg

Notice that the angle in the center of the earth is the same as the angle of depression (A) (since θ is the compliment of the angle of depression and also the compliment of the other angle in the right triangle); and
√[(H+R)^2 - R^2] = √(2RH + H^2)
&
cot(A) = adjacent / opposite = R / √(2RH + H^2)

Answer 7

thats a tough one.lol just kidding