Find distance between two poles.

Question

2 poles, AB of length 2 metres and CD of length 20 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. Find the distance between the two poles.

A) 72m

B) 68m

C) 24m

D) 24.27m

tan(ACB) = AB/BC = 2/77; AB=2. Therefore, we get BC=77. CD is given to be 20 m.

Using Pythagoras formula on BC, BD & CD, I get BD = 74.35m. But the answer given is 72 metres. Can you please tell me where I am going wrong?

Thanks in advance.

Answer 1

Let angle BCD be $\theta$ and let $x$ be the distance between the two poles.

Then,

$tan(\theta) = x/20$ $(1)$

Let $\alpha$ be the angle ACB.

Then,

$tan(\theta+ \alpha) = x/18$

We can evaluate $\alpha$, since $tan(\alpha) = \dfrac{2}{77}$, thus $\alpha = 1.488$

Therefore,

$tan(\theta + 1.488) = x/18$ $(2)$

Equating (1) and (2) gives,

$20 \tan(\theta) = 18tan(\theta + 1.488)$

$20 \tan(\theta) = \dfrac{18(tan\theta+tan1.488)}{1 - tan\theta\; tan1.488}$

Re-arranging gives and setting $tan\theta = \beta$ for simplicity gives,

$0.519\beta^2 - 2\beta + 0.468 = 0$

Solving for $\beta$ gives $\beta = tan(\theta) = 3.6$

And since $tan(\theta) = x/20$, then $x = 20*3.6 = 72$

Answer 2

Let the extension at BD meet AC at O. Required distance is BD.

Let $ OB = y ; BD = x ;\tan \alpha= 2/77, $

We need to solve three equations:

$$ y/x = 1/9 , \tan ( \alpha + \theta )= \dfrac{\tan \alpha+ \tan \theta }{1- \tan \alpha\; \tan \theta }= \dfrac{x+y}{20}, x/20 = \tan \theta ; $$

Eliminate $ y , \theta,$ ( The intermediate missing steps are an exercise).

$$ x\rightarrow 72 $$

If required,

$$ y=8, \tan \theta = 3.6 $$

EDIT1:

First line typo $ AD \rightarrow AC $ corrected.