# Finding the distance of two triangles at .25 degrees

• Jun 4th 2013, 10:07 PM
MetG
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.
Attachment 28530
• Jun 5th 2013, 05:37 AM
MetG
Re: Finding the distance of two triangles at .25 degrees
Any body have an idea how to solve this problem?
• Jun 12th 2013, 10:44 AM
DavidB
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).
• Jun 12th 2013, 12:38 PM
emakarov
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).
• Jun 13th 2013, 07:30 PM
MetG
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
• Jun 14th 2013, 01:34 AM
emakarov
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.).