Answer
Verified
426.6k+ views
Hint- Angles of elevation are given so we find the relation between sides and angles by using some basic trigonometric properties such as $\tan \theta = \dfrac{{{\text{Height}}}}{{{\text{Base}}}}$, To reach the answer we draw the triangle for the given problem.
The pictorial representation of the given problem is shown above.
It is given that the angle of elevation of a tower from a point on the same level as the foot of the tower is ${30^0}$
$\therefore \angle BDA = {30^0}$
Now advancing 150 meter towards the foot of the tower, the angle of elevation of the tower becomes ${60^0}$.
$\therefore \angle BCA = {30^0},\;{\text{DC = 150m}}$
Let the height of tower is h meter
$\therefore AB = h\;{\text{m}}$
Let $BD = x{\text{ m}}$
$ \Rightarrow BC = BD - DC = \left( {x - 150} \right)m$
In triangle ABC
$\tan {60^0} = \dfrac{{{\text{Height}}}}{{{\text{Base}}}} = \dfrac{{AB}}{{BC}} = \dfrac{h}{{x - 150}}$
And we know the value of $\tan {60^0} = \sqrt 3 $
$ \Rightarrow \sqrt 3 = \dfrac{h}{{x - 150}}................\left( 1 \right)$
Now in triangle ABD
$\tan {30^0} = \dfrac{{{\text{Height}}}}{{{\text{Base}}}} = \dfrac{{AB}}{{BD}} = \dfrac{h}{x}$
And we know the value of $\tan {30^0} = \dfrac{1}{{\sqrt 3 }}$
$
\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{h}{x} \\
\Rightarrow x = h\sqrt 3 \\
$
Now substitute the value of x in equation (1) we have
$
\Rightarrow \sqrt 3 = \dfrac{h}{{h\sqrt 3 - 150}} \\
\Rightarrow 3h - 150\sqrt 3 = h \\
\Rightarrow 2h = 150\sqrt 3 \\
\Rightarrow h = 75\sqrt 3 \\
\Rightarrow h = 75 \times 1.732 = 129.9m \\
$
Hence Proved.
Note- whenever we face such types of questions first draw the pictorial representation of the given problem as above then apply the trigonometric identities of tan in triangles which is stated above, then according to given conditions substitute the values and simplify, we will get the required height of the tower.
The pictorial representation of the given problem is shown above.
It is given that the angle of elevation of a tower from a point on the same level as the foot of the tower is ${30^0}$
$\therefore \angle BDA = {30^0}$
Now advancing 150 meter towards the foot of the tower, the angle of elevation of the tower becomes ${60^0}$.
$\therefore \angle BCA = {30^0},\;{\text{DC = 150m}}$
Let the height of tower is h meter
$\therefore AB = h\;{\text{m}}$
Let $BD = x{\text{ m}}$
$ \Rightarrow BC = BD - DC = \left( {x - 150} \right)m$
In triangle ABC
$\tan {60^0} = \dfrac{{{\text{Height}}}}{{{\text{Base}}}} = \dfrac{{AB}}{{BC}} = \dfrac{h}{{x - 150}}$
And we know the value of $\tan {60^0} = \sqrt 3 $
$ \Rightarrow \sqrt 3 = \dfrac{h}{{x - 150}}................\left( 1 \right)$
Now in triangle ABD
$\tan {30^0} = \dfrac{{{\text{Height}}}}{{{\text{Base}}}} = \dfrac{{AB}}{{BD}} = \dfrac{h}{x}$
And we know the value of $\tan {30^0} = \dfrac{1}{{\sqrt 3 }}$
$
\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{h}{x} \\
\Rightarrow x = h\sqrt 3 \\
$
Now substitute the value of x in equation (1) we have
$
\Rightarrow \sqrt 3 = \dfrac{h}{{h\sqrt 3 - 150}} \\
\Rightarrow 3h - 150\sqrt 3 = h \\
\Rightarrow 2h = 150\sqrt 3 \\
\Rightarrow h = 75\sqrt 3 \\
\Rightarrow h = 75 \times 1.732 = 129.9m \\
$
Hence Proved.
Note- whenever we face such types of questions first draw the pictorial representation of the given problem as above then apply the trigonometric identities of tan in triangles which is stated above, then according to given conditions substitute the values and simplify, we will get the required height of the tower.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The cell wall of prokaryotes are made up of a Cellulose class 9 biology CBSE
What organs are located on the left side of your body class 11 biology CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE