Question

$\large \sf \color{purple}Annyeonghaseyo!$
⠀
A tree standing on a horizontal plane is leaning towards east. At two points situated at distances a and b exactly due west on it, the angle elevation of the top are respectively $\alpha$ and $\beta$. Prove that the height of the top from the ground is :
⠀
$\qquad \boxed{ \sf\dfrac{(b - a)tan \alpha \:tan \beta}{tan \alpha - tan \beta} }$

$\large \sf \color{pink}Kamsahamnida ! \: ✿$
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Answer 1

Step-by-step explanation:

$\large \purple{ \bold {annyeong!}}$

Given:

A tree standing on a horizontal plane is leaning towards east. At two points situated at distances a and b exactly due west on it, the angles of elevation of the top are respectively

$\bf\alpha \: and \: \beta$

To Prove:

We have to prove that the height of the top from the ground is

$\bf\frac{(b-a) \tan \alpha \tan \beta}{\tan \alpha-\tan \beta}$

Solution:

Let AB be the tree and C and D are two points situated at distances a and b exactly due weston it.

Draw a perpendicular to the ground from the top of the tree.

From the figure,

$\bf \: \angle BCE=\alpha, \angle BDE=\beta$

Let the height of the perpendicular be h, the distance between A and E be x

$\bf \: In \: right \Delta \: {BCE} \\ \bf {\tan \alpha=\frac{BE}{CE}} \\ \bf =\frac{h}{x+a} \\ \bf \: x+a=\frac{h}{\tan \alpha} \\ \bf \: x=\frac{h}{\tan \alpha}-a...(i)$

Similarly,

$\bf \: In \: right \: \Delta \mathrm{BDE}, \\ \bf\tan \beta=\frac{BE}{DE} \\ \bf \: =\frac{h}{x+b} \\ \bf \: h=(x+b)\tan \beta \\ \bf \: x=\frac{h}{\tan \beta}-b...(ii)$

From (i) and (ii), we get,

$\Rightarrow \bf\frac{h}{\tan \alpha}-a=\frac{h}{\tan \beta}-b \\ \\ \bf\Rightarrow h(\frac{1}{\tan \alpha}-\frac{1}{\tan \beta})=a-b \\ \\ \Rightarrow \bf \: h(\frac{\tan \beta-\tan \alpha}{\tan \alpha \tan\ \beta})=a-b \\ \\ \Rightarrow \bf \: h=\frac{(b-a) \tan \alpha \tan \beta}{\tan \alpha-\tan \beta} \\ \\ \bf \purple{ Hence \: proved.}$