Find the distance from point of observation when angles of depression are unknown

Question

If the two angles of depression are the same how do I determine the distance from the point of observation is?

Answer 1

Let our triangle be $\Delta ABC$, $\measuredangle C=90^{\circ}$, $AD$ be bisector of the triangle,

where $\measuredangle CAD=\measuredangle BAD=\theta$ and $AC=x$.

Thus, by the Pythagoras theorem $$AB=\sqrt{x^2+3.6^2}$$ and since $$\frac{AC}{AB}=\frac{CD}{BD},$$ we obtain: $$\frac{x}{\sqrt{x^2+3.6^2}}=\frac{1.6}{2}$$ or $$x^2=0.64(x^2+3.6^2)$$ or $$0.36x^2=0.64\cdot3.6^2$$ or $$0.6x=0.8\cdot3.6$$ or $$x=4.8.$$

I used the following theorem.

Let $AD$ be bisector of $\Delta ABC$.

Hence, $$\frac{AC}{AB}=\frac{CD}{BD}$$

Answer 2

use that $$\tan(2\theta)=\frac{3.6}{m}$$ and $$\tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}$$

Answer 3

$\require{cancel}$

From $\triangle ABC$, $x=(u+v)\cot2\theta$, similarly, from $\triangle ADC$, $x=u\cot\theta$, hence, \begin{align} (u+v)\cot2\theta&=u\cot\theta ,\\ \cot2\theta\tan\theta &=\frac{u}{u+v} ,\\ \frac{\cos2\theta}{\sin2\theta}\cdot\frac{\sin\theta}{\cos\theta} &=\frac{u}{u+v} ,\\ \frac{\cos2\theta}{2\,\cancel{\sin\theta}\cdot\cos\theta} \cdot\frac{\cancel{\sin\theta}} {\cos\theta} &=\frac{u}{u+v} ,\\ \frac{2\cos^2\theta-1}{2\,\cos^2\theta} &=\frac{u}{u+v} ,\\ 1-\frac{u}{u+v} &= \frac{1}{2\,\cos^2\theta} ,\\ \frac{v}{u+v} &= \frac{1}{2\,\cos^2\theta} ,\\ 2\,\cos^2\theta&=\frac{u+v}{v} ,\\ \cos\theta&=\pm\sqrt{\frac{u+v}{2\,v}} . \end{align}
Note that $\theta<90^\circ$, hence $\cos\theta>0$, we need only positive value of the square root:

\begin{align} \cos\theta&=\sqrt{\frac{u+v}{2\,v}} ,\\ \cot\theta&=\frac{\sqrt{\frac{u+v}{2\,v}}}{\sqrt{1-\frac{u+v}{2\,v}}} =\frac{\sqrt{\frac{u+v}{2\,v}}}{\sqrt{\frac{v-u}{2\,v}}} \\ &=\sqrt{\frac{u+v}{v-u}} ,\\ x&=u\,\sqrt{\frac{u+v}{v-u}} . \end{align}

For $u=1.6,v=2$ we have \begin{align} x&=1.6\,\sqrt{\frac{3.6}{0.4}} =1.6\,\sqrt{9} =4.8\, . \end{align}

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