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# 3D Trig.

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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.

1. Find the height of AB of the building to the nearest metre.
Guest Jun 19, 2014

#1
+1792
+14

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

Rom  Jun 19, 2014
Sort:

#1
+1792
+14

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

Rom  Jun 19, 2014
#2
+85727
+5

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)

CPhill  Jun 19, 2014

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