From the top of a hill h meters high, the angles of depressions of the top and bottom of a pillar are $\alpha $ and $\beta $, respectively. The height (in meters) of the pillar is?
A. $\dfrac{{h[\tan \beta - \tan \alpha ]}}{{\tan \beta }}$
B.$\dfrac{{h[\tan \alpha - \tan \beta ]}}{{\tan \alpha }}$
C. $\dfrac{{h[\tan \alpha + \tan \beta ]}}{{\tan \beta }}$
D. $\dfrac{{h[\tan \alpha + \tan \beta ]}}{{\tan \alpha }}$
Answer
300.6k+ views
Hint: Since the problem is based on applications of trigonometry hence, it is necessary to draw the figure for better visualization of the problem. Here, we will use proper trigonometric identities and ratios and then equate them to get the required dimension of a pillar and find out the solution to a given problem.
Complete step by step Solution:
The angles of depression of the top and bottom of a pillar $\alpha $ and $\beta $ respectively (given).
Let the height of the hill $(AD)$ be $h$ and the height of the pillar $(CE)$ be ${h_1}$ as shown in the figure.
Let the distance between the feet of the hill and pillar $(DE)$ be $x$.

In \[\vartriangle ABC\],
$\tan \alpha = \dfrac{P}{B} = \dfrac{{AB}}{{BC}}$
From the figure, it is clear that $AB = h - {h_1}$ and $BC = x$, substitute these values in the above expression, we get
$\tan \alpha = \dfrac{{h - {h_1}}}{x}$
$x = \dfrac{{h - {h_1}}}{{\tan \alpha }}$ …. (1)
In \[\vartriangle ADE\],
$\tan \beta = \dfrac{P}{B} = \dfrac{{AD}}{{DE}}$
From the figure, it is clear that $AD = h$ and $DE = x$, substitute these values in the above expression, we get
$\tan \beta = \dfrac{h}{x}$
$x = \dfrac{h}{{\tan \beta }}$ …. (2)
Equating eq. (1) and (2), we get
$\dfrac{{h - {h_1}}}{{\tan \alpha }} = \dfrac{h}{{\tan \beta }}$
By Cross-multiplication,
$h.\tan \beta - {h_1}.\tan \beta = h.\tan \alpha $
$h.\tan \beta - h.\tan \alpha = {h_1}.\tan \beta $
Taking common$h$from the above expression, we get
$h(\tan \beta - \tan \alpha ) = {h_1}.\tan \beta $
${h_1} = \dfrac{{h(\tan \beta - \tan \alpha )}}{{\tan \beta }}$
Thus, the Height of the pillar is $\dfrac{{h(\tan \beta - \tan \alpha )}}{{\tan \beta }}$ .
Hence, the correct option is A.
Note: In the problems based on the application of trigonometry, apply the required trigonometric ratios to get the accurate solution to the problem. Sometimes, the problem seems to be complex while analyzing the problem hence, it is advised to draw a figure and calculate ratios one by one carefully.
Complete step by step Solution:
The angles of depression of the top and bottom of a pillar $\alpha $ and $\beta $ respectively (given).
Let the height of the hill $(AD)$ be $h$ and the height of the pillar $(CE)$ be ${h_1}$ as shown in the figure.
Let the distance between the feet of the hill and pillar $(DE)$ be $x$.

In \[\vartriangle ABC\],
$\tan \alpha = \dfrac{P}{B} = \dfrac{{AB}}{{BC}}$
From the figure, it is clear that $AB = h - {h_1}$ and $BC = x$, substitute these values in the above expression, we get
$\tan \alpha = \dfrac{{h - {h_1}}}{x}$
$x = \dfrac{{h - {h_1}}}{{\tan \alpha }}$ …. (1)
In \[\vartriangle ADE\],
$\tan \beta = \dfrac{P}{B} = \dfrac{{AD}}{{DE}}$
From the figure, it is clear that $AD = h$ and $DE = x$, substitute these values in the above expression, we get
$\tan \beta = \dfrac{h}{x}$
$x = \dfrac{h}{{\tan \beta }}$ …. (2)
Equating eq. (1) and (2), we get
$\dfrac{{h - {h_1}}}{{\tan \alpha }} = \dfrac{h}{{\tan \beta }}$
By Cross-multiplication,
$h.\tan \beta - {h_1}.\tan \beta = h.\tan \alpha $
$h.\tan \beta - h.\tan \alpha = {h_1}.\tan \beta $
Taking common$h$from the above expression, we get
$h(\tan \beta - \tan \alpha ) = {h_1}.\tan \beta $
${h_1} = \dfrac{{h(\tan \beta - \tan \alpha )}}{{\tan \beta }}$
Thus, the Height of the pillar is $\dfrac{{h(\tan \beta - \tan \alpha )}}{{\tan \beta }}$ .
Hence, the correct option is A.
Note: In the problems based on the application of trigonometry, apply the required trigonometric ratios to get the accurate solution to the problem. Sometimes, the problem seems to be complex while analyzing the problem hence, it is advised to draw a figure and calculate ratios one by one carefully.
Recently Updated Pages
The HCF of two numbers is 96 and their LCM is 1296 class 10 maths JEE_Main

If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

Four persons A B C and D initially at the corners of class 11 physics JEE_Main

If a parabola whose length of latus rectum is 4a touches class 11 maths JEE_Main

Which of the following are correct regarding the normal class 13 maths JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

Trending doubts
JEE Main Marks vs Percentile 2026: Predict Your Score Easily

JEE Main Cutoff 2026: Category-wise Qualifying Percentile

JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Marks vs Rank 2026: Expected Rank for 300 to 0 Marks

NIT Cutoff 2026: Tier-Wise Opening and Closing Ranks for B.Tech. Admission

JEE Mains 2027 Subject Wise Percentile Explained

Other Pages
CBSE Class 10 Maths Question Paper 2026 OUT Download PDF with Solutions

Complete List of Class 10 Maths Formulas (Chapterwise)

NCERT Solutions For Class 10 Maths Chapter 11 Areas Related To Circles - 2026-27 Free PDF Download (Login Required)

All Mensuration Formulas with Examples and Quick Revision

NCERT Solutions For Class 10 Maths Chapter 13 Statistics - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 10 Maths Chapter 14 Probability - 2026-27 Free PDF Download (Sign-in Required)

