Let AB and PQ be two vertical poles, 160m apart from each other.

Question

Let $AB$ and $PQ$ be two vertical poles, 160m apart from each other. Let $C$ be the middle point of $B$ and $Q$, which are feet of these two poles. Let $\dfrac{\pi}8$ and $	heta$ be the angles of elevation from $C$ to $P$ and $A$, respectively. If the height of pole $PQ$ is twice the height of pole $AB$, then $	an2	heta$ is equal to:

Accepted Answer

Let $AB=l$, so $PQ=2l$; $C$ is midpoint of $BQ$, so $BC=CQ=80$.

$$\tan\theta = \frac l{80} \quad \text{(i)}, \qquad \tan\frac{\pi}8 = \frac{2l}{80} \quad \text{(ii)}$$

From (i) and (ii): $\dfrac{\tan\theta}{\tan\frac{\pi}8}=\dfrac12 \implies \tan\theta = \dfrac12\tan\dfrac{\pi}8 = \dfrac{\sqrt2-1}2$

$$\tan2\theta = \frac{2\tan\theta}{1-\tan^2\theta} = \frac{\sqrt2-1}{1-\left(\frac{\sqrt2-1}2\right)^2} = \frac{3-2\sqrt2}4$$

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