Answer
Verified
375.5k+ views
Hint: The first thing you need to do is to draw a neat diagram. Then use the trigonometric ratios to find the height of the towers.
Complete step-by-step answer:
To start with the question, let us first draw a neat diagram of the situation given in the figure.
Now it is given the question that CD = 5m, and we let CB to be h. Also, let AB to be b.
It is also given that $\angle DAB=60{}^\circ $ and $\angle ACO=45{}^\circ $ .
Also, it is clear from the diagram that $\angle BCO=90{}^\circ $ .
$\therefore \angle BCO=\angle ACO+\angle BCA=90{}^\circ $
Now if we put the value of $\angle ACO$ , we get
$45{}^\circ +\angle BCA=90{}^\circ $
$\Rightarrow \angle BCA=45{}^\circ $
Now let us focus on the right-angled triangle ADB.
We know $\tan \theta =\dfrac{perpendicular}{base}$ .So, we can say that: $\tan \left( \angle DAB \right)=\dfrac{BD}{AB}=\dfrac{h+5}{b}$ .
$\therefore \tan 60{}^\circ =\dfrac{h+5}{b}$
We also know that $\tan 60{}^\circ =\sqrt{3}$ , so putting the value in our equation, we get
$\sqrt{3}=\dfrac{h+5}{b}$
$\Rightarrow \sqrt{3}b=h+5$
$\Rightarrow b=\dfrac{h+5}{\sqrt{3}}$
Now we shall focus on the right-angled triangle ABC.
$\tan \left( \angle BCA \right)=\dfrac{AB}{CB}=\dfrac{b}{h}$
On putting $\angle BCA=45{}^\circ $, we get
$\tan \left( 45{}^\circ \right)=\dfrac{b}{h}$
We know that the value of $\tan 45{}^\circ $ is equal to 1. So, our equation becomes:
$1=\dfrac{b}{h}$
$\Rightarrow h=b$
Now we will substitute the value of b in terms of h from equation (i). On doing so, our equation becomes:
$h=\dfrac{h+5}{\sqrt{3}}$
$\sqrt{3}h-h=5$
As given in the question we put the value of $\sqrt{3}$ as 1.732, we get
$1.732h-h=5$
$\Rightarrow .732h=5$
$\Rightarrow h=\dfrac{5}{.732}=6.83$
So, we can conclude that the height of the tower is 6.83 m.
Note: While solving such questions, make sure that your diagram is correct and satisfies all the conditions given in the question, as the key to such questions is the diagram and the knowledge of simple trigonometric ratios.
Complete step-by-step answer:
To start with the question, let us first draw a neat diagram of the situation given in the figure.
Now it is given the question that CD = 5m, and we let CB to be h. Also, let AB to be b.
It is also given that $\angle DAB=60{}^\circ $ and $\angle ACO=45{}^\circ $ .
Also, it is clear from the diagram that $\angle BCO=90{}^\circ $ .
$\therefore \angle BCO=\angle ACO+\angle BCA=90{}^\circ $
Now if we put the value of $\angle ACO$ , we get
$45{}^\circ +\angle BCA=90{}^\circ $
$\Rightarrow \angle BCA=45{}^\circ $
Now let us focus on the right-angled triangle ADB.
We know $\tan \theta =\dfrac{perpendicular}{base}$ .So, we can say that: $\tan \left( \angle DAB \right)=\dfrac{BD}{AB}=\dfrac{h+5}{b}$ .
$\therefore \tan 60{}^\circ =\dfrac{h+5}{b}$
We also know that $\tan 60{}^\circ =\sqrt{3}$ , so putting the value in our equation, we get
$\sqrt{3}=\dfrac{h+5}{b}$
$\Rightarrow \sqrt{3}b=h+5$
$\Rightarrow b=\dfrac{h+5}{\sqrt{3}}$
Now we shall focus on the right-angled triangle ABC.
$\tan \left( \angle BCA \right)=\dfrac{AB}{CB}=\dfrac{b}{h}$
On putting $\angle BCA=45{}^\circ $, we get
$\tan \left( 45{}^\circ \right)=\dfrac{b}{h}$
We know that the value of $\tan 45{}^\circ $ is equal to 1. So, our equation becomes:
$1=\dfrac{b}{h}$
$\Rightarrow h=b$
Now we will substitute the value of b in terms of h from equation (i). On doing so, our equation becomes:
$h=\dfrac{h+5}{\sqrt{3}}$
$\sqrt{3}h-h=5$
As given in the question we put the value of $\sqrt{3}$ as 1.732, we get
$1.732h-h=5$
$\Rightarrow .732h=5$
$\Rightarrow h=\dfrac{5}{.732}=6.83$
So, we can conclude that the height of the tower is 6.83 m.
Note: While solving such questions, make sure that your diagram is correct and satisfies all the conditions given in the question, as the key to such questions is the diagram and the knowledge of simple trigonometric ratios.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
A group of fish is known as class 7 english CBSE
The highest dam in India is A Bhakra dam B Tehri dam class 10 social science CBSE
Write all prime numbers between 80 and 100 class 8 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Onam is the main festival of which state A Karnataka class 7 social science CBSE
Who administers the oath of office to the President class 10 social science CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Kolkata port is situated on the banks of river A Ganga class 9 social science CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE