Find distance between two poles. 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.
 A: 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 $x = 20*3.6 = 72$
A: 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.
