Question

Two poles, AB of length two metres and CD of length twenty metres are erected vertically with bases at B and D. The two poles are at a distance not less than twenty metres. It is observed that \tan angle ACB = 2/77. The distance between the two poles is

Select one:
a. 72 m
b. 24.27 m
c. 68 m
d. 80 m
e. 24 m​

Answer 1

Answer:

72

Step-by-step explanation:

Let angle BCD be θ and let x be the distance between the two poles.

Then,

tan(θ)=x/20 (1)

Let α be the angle ACB.

Then,

tan(θ+α)=x/18

We can evaluate α, since tan(α)=277, thus α=1.488

Therefore,

tan(θ+1.488)=x/18 (2)

Equating (1) and (2) gives,

20tan(θ)=18tan(θ+1.488)

20tan(θ)=18(tanθ+tan1.488)1−tanθtan1.488

Re-arranging gives and setting tanθ=β for simplicity gives,

0.519β2−2β+0.468=0

Solving for β gives β=tan(θ)=3.6

And since tan(θ)=x/20, then x=20∗3.6=72

Answer 2

Answer:

d.80 m

Step-by-step explanation:

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