David Gwyn J. answered • 11/02/20
Highly Experienced Tutor (Oxbridge graduate and former tech CEO)
This is a problem with trigonometry, and we need to know about right angle triangles and sines/cosines/tangents. Plus, a bit of simultaneous equations in 2 unknowns.
We need to find H, the height of the tower.
The tower is assumed to be vertical, so this gives us our right angle triangle. The height H is the opposite side from the angle of elevation, while the adjacent side is the distance from the tower. The hypotenuse is the line of sight to the top of the tower, but we don't need this for this problem.
From point B, we can say that the tower is x feet distance away
tan b = opposite/adjacent = H/x
=> H = x tan b
and we are given angle b is 16.1 degrees, so H = x tan (16.1 deg)
Similarly, from point A, we can say the tower is x + 35.9 feet away
tan a = H / (x + 35.9)
=> H = (x + 35.9) tan a
we are also given angle a which is 12.3 degrees, so H = (x + 35.9) tan (12.3 deg)
At this point we have our two simultaneous equations, in two unknowns H and x. But as they are both equal to H, we can set them equal to each other to immediately eliminate H. Hence,
x tan (16.1 deg) = (x + 35.9) tan (12.3 deg)
rearrange to get x
=> x tan (16.1 deg) = x tan (12.3) + 35.9 tan (12.3 deg)
=> x tan (16.1 deg) - x tan (12.3 deg) = 35.9 tan (12.3 deg)
=> x ( tan (16.1 deg) - tan (12.3 deg) ) = 35.9 tan (12.3 deg)
=> x = 35.9 tan (12.3 deg) / ( tan (16.1 deg) - tan (12.3 deg) )
=> 110.87 or 111 feet (to nearest foot)
And as H = x tan (16.1 deg) then H = 110.87 tan (16.1 deg) = 32 feet (to nearest foot).
As usual, it's best to double-check our solution. If height = 32, and distance to B is 111, then distance to A is 147 feet. Our tan here is 32/147, so we need arctan (32/147) = 12.3 degrees, as required. There are also online angle of elevation calculators so you can use one to check we're correct.
