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?

Answer 1

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.

Answer 2

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?

Answer 3

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.