The angle of elevation of the top A of a building from a point C due south of it is 25 degrees. At a second point D, which is 160 metres due west of C, the angle of elevation of the top of the building is 20 degrees. Point B is the bottom of the vertical building and on the same horizontal plane as D and C.
+6248 Let C be the point (0,0,0)
Then D is (-160,0,0)
B is (0,y,0)
A is (0,y,z)
From what we are given
$$\tan(25deg) = \dfrac z y$$
$$\tan(20deg) = \dfrac{z}{\sqrt{160^2+y^2}}$$
solving this we get
z=93.16m which is the the length of AB, i.e. the height of the building
+128408 Using the information in the first part of the problem, let d be the distance from C to B and h be the height of the building. So we have,
tan(25) = h/d Solving for h we have, h = d*tan(25)
And, since we have a right triangle with two legs d and 160, the distnce that D is from B = √(1602 + d2)
So we have
tan(20) = (h)/√(1602 + d2) = (d*tan(25)) / √(1602 + d2)
Simpifying, we have
tan(20)/ tan(25) = d/√(1602 + d2) and we can write
tan(25)/tan(20) = √(1602 + d2) /d Square both sides
[tan(25)/tan(20)]2 = (1602 + d2)/d2 and the right side = 160/d2 + 1 Subtract 1 from both sides
([tan(25)/tan(20)]2 - 1) = 1602/d2 and we can write
d2 = 1602 / ([tan(25)/tan(20)]2 - 1) Simplifying this, we have
d2 = 1602 / (.6413959572703699169) Now....take the square root of both sides
d = 199.7822378461935026188794268 m
And using .... h = d*tan(25) ....we have....
h = (199.7822378461935026188794268)* tan(25) ≈ 93.159 m or just 93 m
I think that's it.....I haven't worked one like this before....thanks for submitting it....I hope I haven't made any grevious mistakes !! (P.S. ....thanks to alan for catching my previous error.....I went back and corrected it so that now, it appears that I'm smarter than I might really be....also...rom's method is better and actually conforms to what was asked....but I guess my approach shows that there's more than one way to skin a cat !!) (so to speak)
