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Hint : (Make figure , analyze from it , get the problem solved)

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 .

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 .