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Some Applications of Trigonometry — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsSome Applications of Trigonometry5 Marks
Question
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 $\alpha$ and $\beta.$ Prove that the height of the top from the ground is, $\frac{(\text{b}-\text{a})\tan\alpha\tan\beta}{\tan\alpha-\tan\beta}.$
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Answer

Let CD is the tree which is leaning towards east and A and B are two points on the West making angles of elevation with top C of the tree as $\alpha$ and $\beta.$
A and B are at the distance of a and b from the foot of the tree CD, then AD = a, BD = b.

Draw $\text{CL}\bot\text{BD}$ produced and let DL = x and CL = h
Now in right $\triangle\text{CAL,}$
$\tan\alpha=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{CL}}{\text{AL}}=\frac{\text{h}}{\text{a}+\text{x}}$
$\Rightarrow \text{a}+\text{x}=\frac{\text{h}}{\tan\alpha}\ .....(\text{i})$
Similarly in right $\triangle\text{CBL,}$
$\tan\beta=\frac{\text{CL}}{\text{BL}}=\frac{\text{h}}{\text{b}+\text{x}}$
$\Rightarrow\ \text{b}+\text{x}=\frac{\text{h}}{\tan\beta}\ .....(\text{ii})$
From (i) and (ii)
$\text{x}=\frac{\text{h}}{\tan\alpha}-\text{a}$
$\text{x}=\frac{\text{h}}{\tan\beta}-\text{b}$
$\therefore\ \frac{\text{h}}{\tan\alpha}-\text{a}=\frac{\text{h}}{\tan\beta}-\text{b}$
$\Rightarrow\ \frac{\text{h}}{\tan\beta}-\frac{\text{h}}{\tan\alpha}=\text{b}-\text{a}$
$\Rightarrow\ \text{h}\Big(\frac{1}{\tan\beta}-\frac{1}{\tan\alpha}\Big)=\text{b}-\text{a}$
$\Rightarrow\ \text{h}\Big(\frac{\tan\alpha-\tan\beta}{\tan\beta\tan\alpha}\Big)=\text{b}-\text{a}$
$\Rightarrow\ \text{h}=\frac{(\text{b}-\text{a})\tan\alpha\tan\beta}{\tan\alpha-\tan\beta}.$
Hence proved.

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