From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects

Name: From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects
Uploaded: Mon, 2021-02-01 18:00
Description: From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects

#Maths
#Introduction to Trigonometry
#NCERT
#Exemplar Maths for Class 10

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#8.2
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#8.1
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From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects.

Answers (1)

According to question

Let x and y be two objects and $\beta$ and $\alpha$ use the angles of depression of two objects.
In $\bigtriangleup$ AOX
$\tan \beta = \frac{h}{Ox}$ $\because \tan \theta = \frac{Perpendicular}{Base}$
$Ox= \frac{h}{\tan \beta }$
$Ox= h\cot \beta \cdots \left ( 1 \right )$
In $\bigtriangleup$ AOY
$\tan \alpha = \frac{h}{Oy}= \frac{h}{Ox+xy}\; \; \left ( \because Oy= Ox+xy \right )$
$Ox+xy= \frac{h}{\tan \alpha }$
$Ox+xy= h\cot \alpha$
$xy= h\cot \alpha- Ox$
$xy= h\cot \alpha- h\cot \beta$ (from equation (1))
$xy= h\left ( \cot \alpha- \cot \beta \right )$
Hence the distance between two objects is