Question

A vertical pole 35m high, standing on sloping ground is braced by a wire which extends from the top of the pole to a point on the ground 25 m from the foot of the pole. If the pole subtends an angle of 30 degrees at the point where the wire reaches the ground, how long is the wire?

Answers

as usual, draw a diagram. Let
T be the top of the pole
B be the bottom of the pole
P be the point where the wire meets the ground.

Draw a horizontal line from P, and extend TB into the ground, so that
Q is the intersection of the horizontal from P and the vertical from T

Then if we let
x be the horizontal distance PQ
y be the vertical distance QB
z be the length of the wire PT
θ be the angle the road makes with the horizontal
we have


y/x = tanθ
(y+35)/x = tan(θ+30°)
So,

tan(θ+30°) = (tanθ + 1/√3)/(1 - tanθ/√3)
(y/x + 1/√3)/(1 - y/x√3) = (x+35)/y
x^2+y^2 = 25^2
That gives us
x = 18.44 and y=16.88

z^2 = x^2 + (y+35)^2
z^2 = 18.44^2 + (16.88+35)^2
z = 55.06

So, the wire is about 55m long.
My bad - switched x and y:

(y/x + 1/√3)/(1 - y/x√3) = (y+35)/x
x^2+y^2 = 25^2
x = 19.408, y=15.759
so, z = 54.343

not very far off, really, eh?
it s laborious using a right triangle theorem, however if sine and cosine law for oblique triangle were discussed by the instructor, it s an easy
work.

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