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 base of the mountain be at the x-y-z origin (0,0,0)
let the mountain be of height h, then the top is at (0,0,h)
now assume the surveyor is at (x,y,0) when taking the first measurment
and (x+15,y,0) when taking the second (east being the positive x direction)
now for the first measurement we have (angle in degrees)
tan(5)=h/sqrt(x^2+y^2)
and from the second
tan(6)=h/sqrt((x+15)^2+y^2)
and the third
tan(4)=h/sqrt((x+25)^2+y^2)
so now we have 3 variables and 3 equations, so we can solve for h,x,y
rearranging and squaring we get
x^2+y^2=h^2/tan^2(5)
x^2+30x+225+y^2=h^2/tan^2(6)
x^2+50x+625+y^2=h^2/tan^2(4)
subtracting the first from the second and third gives us
30x+225=h^2(1/tan^2(6)-1/tan^2(5))
50x+625=h^2(1/tan^2(4)-1/tan^2(6))
multiply the first by 5 and second by 3
150x+1125=5h^2(1/tan^2(6)-1/tan^2(5))
150x+1875=3h^2(1/tan^2(4)-1/tan^2(6))
subtracting gives us
750=h^2(3/tan^2(4)-3/tan^2(6)-5/tan^2(6)+5/tan^2(5))
this then gives us
h=1.175714....
so the mountain is approximately 1.17 miles high, or 6177 feet
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Posted by Daniel
on 2019-07-31 07:54:03 |
solution
