A surveyor on a flat plain sees a mountain in the distance. The angle of elevation to the peak is 5 degrees. The surveyor drives 15 miles east. From there, the angle of elevation is 6 degrees. Driving an additional 10 miles east, the angle of elevation is 4 degrees. How high is the mountain?
Let the location of the mountain be at (0,0,0).
The height of the mountain is m miles(say).
When taking the first measurement, the surveyor is at (x, y, 0) (say)
-> Then, at the time of the second measurement, the surveyor is at (x+15, y, 0)
-> At the time of the third measurement, the surveyor is at (x+25, y, 0)
Accordingly, we must have:
m
-------------------- = tan(5)
V(x^2+y^2)
m
------------------------------ = tan(6)
V{(x+15)^2 + y^2}
m
------------------------------ = tan(4)
V{(x+25)^2+y^2}
Thus,
m^2
x^2+y^2 = ------------------
tan^2(5)
m^2
x^2 +30x +225 + y^2 = -----------------------
tan^2(6)
m^2
x^2 + 50x+625 + y^2 = --------------------
tan^2(4)
So, we have:
1 1
30x+225 = m^2 { -------------- - -------------- }
tan^2(6) tan^2(5)
1 1
20x + 400 = m^2 { -------------- - --------------}
tan^2(4) tan^2(6)
Multiplying the second equation by 3 and subtracting twice the first equation from it, we must have:
3 3 2 2
750 = m^2 { --------------- - --------------- - --------------- + --------------- }
tan^2(4) tan^2(6) tan^2(6) tan^2(5)
3 2 5
= m^2 { ---------------- + ------------- - ------------ }
tan^2(4) tan^2(5) tan^2(6)
~= m^2 * 422.203702606426
=> m^2 ~= 1.77693706094857
=> m ~= 1.33281420539206
Consequently, the required height of the mountain is approximately 1.3328142 miles which is nearly 1.3328142*5280 = 7037.259 feet.
Edited on July 15, 2022, 2:20 am