SOLUTION: Hi, can you please help me solve this problem? Thank you. A surveyor wishes to find the distance between two inaccessible points A and B. As shown in the figure, two points C and

Question 1207648: Hi, can you please help me solve this problem? Thank you.
A surveyor wishes to find the distance between two inaccessible points A and B. As shown in the figure, two points C and D are selected from which it is possible to view both A and B. The distance CD and the angles ACD, ACB, BDC, and BDA are then measured. If CD=120~ft, ∠ACD=115°, ∠ACB=92°, ∠BDC=125°, and ∠BDA=100°, approximate the distance AB.
Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!

Here's the drawing, approximately but not exactly to scale. This is a law of sines and cosines problem. I'm not going to finish it for you. These aren't special angles, so you're going to have a lots of long decimals. The more you round off, the less accurate your answer will be. By subtracting I found the two angles in green. You have angle-side-angle in triangle CED, so use the law of sines to solve for the other 3 parts of triangle CED. Then, by subtracting angles find the two acute angles at E. Then you will have angle-side-angle in each of the triangles ACE and BED, so use the law of sines on triangle ACE to find AE, then again on triangle BED to find BE. Then for triangle AEB, you will have side-angle-side. That will be AE, angle AEB (same as angle CED), and BE. and you can then find AB, using the law of cosines on triangle AEB. Happy solving! Edwin

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Refer to the diagram tutor Edwin has drawn.

There are a lot of triangles to keep track of, so it might be helpful to peel the triangles ACD and BCD apart to get this

The diagrams aren't to scale.

Recall that for any triangle, the interior angles always add to 180 degrees.
We'll use this fact to find the missing angles of triangles ACD and BCD.

For triangle ACD, the missing angle is A = 180-C-D = 180-115-25 = 40 degrees.
For triangle BCD, the missing angle is B = 180-C-D = 180-23-125 = 32 degrees.

Let's update the diagram

From here, we have a few pathways we could take.
The path I'll take is to determine the side AC (from triangle ACD) and determine side BC (from triangle BCD).
Use the Law of Sines to find these lengths.

I'll get you started with the setup equations
AC/sin(D) = CD/sin(A)
AC/sin(25) = 120/sin(40)
and
BC/sin(D) = CD/sin(B)
BC/sin(125) = 120/sin(32)
Make sure that your calculator is set to degrees mode.
I'll let the student finish this subsection.

--------------------------------------------------------------------------------

Let's focus on triangle ABC.

The last batch of steps would be to use the Law of Cosines to find the length of side AB.
x = length of AC
y = length of BC
z = length of AB

Law of Cosines
z^2 = x^2 + y^2 - 2*x*y*cos(C)
z^2 = x^2 + y^2 - 2*x*y*cos(92)

I'll let the student finish up.

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