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Finding the distance of two triangles at .25 degrees

Hi I needed some help with some trig. I need to find the distance of both triangles A and B at a .25 degrees, and then find the distance of AB. If some one can help me out step by step with the formula on how to accomplish this would be most appreciated thanks you.

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Re: Finding the distance of two triangles at .25 degrees

Any body have an idea how to solve this problem?

Re: Finding the distance of two triangles at .25 degrees

Are you sure that is supposed to be 0.25 degree? Not radians?

You could start by calculating the difference in height between A and B and create a small triangle for the difference: A - B = 0.2325 - 0.21 = 0.0225.

Since the angle is the same, a, you know tan(a) = 0.0225/cos(a).

Re: Finding the distance of two triangles at .25 degrees

Quote:

Originally Posted by

**DavidB** Since the angle is the same, a, you know tan(a) = 0.0225/cos(a).

This is an approximate formula, isn't it? And why involve cos(a)?

AB = (A - B)cot(a).

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Re: Finding the distance of two triangles at .25 degrees

DavidB/emokarov thank you for the the response was trying to work out the problem but got 80 percent complete and got stuck. Can you look at my problem and see what am doing wrong? I've exhausted the attempt to find the distance of AB. thanks

Re: Finding the distance of two triangles at .25 degrees

There is no need to use the right triangle calculator to find 48.128. You can find cot(0.25) = 1/tan(0.25) using a regular calculator. Anyway, you are correct, cot(0.25) = 229.18. Then, as I said in post #4, AB = (A - B)cot(a) = 0.0225 * 229.18 = 5.16 (rounding to 2 d.p.).