SOLUTION: To measure the height of a mountain, a surveyor takes two sightings of a peak 900 meters apart on a direct line to the mountain. The two different angles of elevation are, 47 and

Question 1198061: To measure the height of a mountain, a surveyor takes two sightings of a peak 900 meters apart on a direct line to the mountain. The two different angles of elevation are, 47 and 35 degrees. If the transit is 2 meters high, how tall is the mountain?
One of the following is the answer. Which one?
A) 1660 meters
B) 1752 meters
C) 1818 meters
D) 2059 meters
E) 2278 meters
Found 2 solutions by Alan3354, math_tutor2020:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
See
Angles/1197859:
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Same problem, different numbers.
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

This is one way to do the drawing

The observation points are at B and C.
The mountain is represented by segment FD, which has a height of h+2.

Segments:
AB = x
BC = 900
FA = h
AD = CE = 2
FD = h+2

Angles:
angle ABF = 47 degrees
angle ACF = 35 degrees

Let
u = tan(47)
v = tan(35)
which will help us do the symbolic operations shown below a bit easier.

Focus on triangle ABF.
Use the tangent ratio to say
tan(angle) = opposite/adjacent
tan(B) = FA/AB
tan(47) = h/x
u = h/x
ux = h
h = ux

Now focus on triangle ACF
tan(C) = FA/AC
tan(35) = h/(x+900)
v = h/(x+900)
v(x+900) = h
vx + 900v = h
vx + 900v = ux
900v = ux - vx
900v = (u-v)x
x = 900v/(u-v)

Therefore,
h = ux
h = u*900v/(u-v)
h = 900uv/(u-v)
we have this fairly tidy formula to calculate the height based on the tangents of the angles mentioned, and also based on the distance between the observation points from B to C.

Now plug in the definitions for u and v mentioned earlier
h = 900uv/(u-v)
h = 900*tan(47)*tan(35)/(tan(47)-tan(35))
h = 1815.86000966682

which leads to:
h+2 = 1815.86000966682+2 = 1817.86000966681

When rounding to the nearest whole number we get 1818 meters as the final answer.

Related Question
https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1198040.html

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