How Do You Calculate the Distance to a Mountain Using Trigonometry?

[ataglance05]

A surveyor is trying to determine the height of a mountain. First, he msut determine how far away it is. He establishes a base line of 1km and measures the angle to the summit from both ends of the base line. The angle on the right side is 88degrees and the angle on the left end is 88degrees. (Mountain is centered on the base line.)

How far away is the mountain? (what is the perpendicular distance from the base line to the mountain?

is this a trick question or is it just simply 1000meters?

 

[Stevedye56]

You want the Y value of the triangle if you think of it on a unit circle in quadrant 1. You are 1000meters away call it "x". It is asking you how tall the mountain is vertically not how far away it is horizontally.

 

[ataglance05]

The surveyor laid the baseline at some distance x from the mountain. Since the triangle formed is isosceles, the perpendicular from the apex (x) bisects the side. So x = 500 tan 88.
However, this is the distance to the mountain summit, not the ground distance from the base line to the mountain. So this may not be correct...
http://answerboard.cramster.com/Answer-Board/Image/2007652157276331667744716200003866.jpg

so is this now what i must solve?? is 88 degrees in the right area? is that even the angle?:[

 

[Stevedye56]

Hey ataglance05. I think I read the problem wrong, I thought that the x distance was 1000m and there was only one triangle to be drawn. My mistake, however I think you should be able to solve the problem using the law of sines.

[tex] \frac {a} {sin(A)} = \frac {b} {sin(B)} = \frac {c} {sin(C)} [/tex]

Edit: Your first link shows a correct diagram.