The angle of elevation of the top of a tower from a point A on the ground is ${30^ \circ }$. On moving a distance of $20$ metres towards the foot of the tower to a point B the angle of elevation increases to ${60^ \circ }$. Find the height of the tower and the distance of the tower from point A.
Last updated date: 25th Mar 2023
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Answer
209.8k+ views
Hint : (Make figure , analyze from it , get the problem solved)
Complete step-by-step answer:
From the figure
AB$ = 20m$ (Given)
Given
$
\angle CAB = {30^ \circ } \\
\angle CBD = {60^ \circ } \\
$
To find CD, The height of the tower
First we will consider triangle ACD
We know ,
$
\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }} = \dfrac{{CD}}{{AD}} \\
AD = \sqrt 3 \,CD\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......({\text{i}}) \\
$
In triangle CBD we have
$
\tan {60^ \circ } = \dfrac{{CD}}{{BD}} = \sqrt 3 \\
\dfrac{{CD}}{{\sqrt 3 }} = \,BD\, \\
\\
AB = AD - BD \\
20 = \sqrt 3 \,CD - \,\,\dfrac{{CD}}{{\sqrt 3 }} = CD\,\left( {\dfrac{2}{{\sqrt 3 }}} \right)\, \\
\therefore \,\,CD = 10\sqrt 3 \\
$
Hence the height of the tower is $10\sqrt 3 = 17.34\,\,m$
Now we have to find the Length AD
So, Again we can consider the triangle ADC
$
tan{30^ \circ } = \dfrac{1}{{\sqrt 3 }} = \dfrac{{10\sqrt 3 }}{{AD}} \\
AD = 10\sqrt 3 .\sqrt 3 = 30\,\,m \\
$
Hence the distance of the tower from point A is 30 m.
Note :- Whenever these types of questions arise you must have to use the trigonometric concepts for triangles . Using the values of respective angles you can simply find any length present in the figure using “tan” in most of the cases will make your problem solved easily .
Given
$
\angle CAB = {30^ \circ } \\
\angle CBD = {60^ \circ } \\
$
To find CD, The height of the tower
First we will consider triangle ACD
We know ,
$
\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }} = \dfrac{{CD}}{{AD}} \\
AD = \sqrt 3 \,CD\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......({\text{i}}) \\
$
In triangle CBD we have
$
\tan {60^ \circ } = \dfrac{{CD}}{{BD}} = \sqrt 3 \\
\dfrac{{CD}}{{\sqrt 3 }} = \,BD\, \\
\\
AB = AD - BD \\
20 = \sqrt 3 \,CD - \,\,\dfrac{{CD}}{{\sqrt 3 }} = CD\,\left( {\dfrac{2}{{\sqrt 3 }}} \right)\, \\
\therefore \,\,CD = 10\sqrt 3 \\
$
Hence the height of the tower is $10\sqrt 3 = 17.34\,\,m$
Now we have to find the Length AD
So, Again we can consider the triangle ADC
$
tan{30^ \circ } = \dfrac{1}{{\sqrt 3 }} = \dfrac{{10\sqrt 3 }}{{AD}} \\
AD = 10\sqrt 3 .\sqrt 3 = 30\,\,m \\
$
Hence the distance of the tower from point A is 30 m.
Note :- Whenever these types of questions arise you must have to use the trigonometric concepts for triangles . Using the values of respective angles you can simply find any length present in the figure using “tan” in most of the cases will make your problem solved easily .
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