Answer
405.6k+ views
Hint: Draw a diagram by carefully reading the question and applying the constraints given in it. Assume two variables (say ‘x’ and ‘y’) for distance of point ‘A’ and ‘B’ respectively from the base of the tower. Then use the formula $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ in two triangle separately to find ‘x’ and ;y; and then calculate distance between ‘A’ and ‘B’ by substituting ‘x’ from ‘y’.
Complete step by step answer:
According to question, we can draw the following diagram-
We have to find the distance between points ‘A’ and ‘B’. Let us assume the distance of point ‘A’ from the base of tower to be x.m and the distance of point B from the base of tower to be ‘y’m.
From the diagram, we can see that-
Distance between points ‘A’ and ‘B’ $=\left( y-x \right)m$ …………………………….. (1)
To find the distance between ‘A’ and ‘B’, we need to find ‘x’ and ‘y’.
Let us find ‘x’ first-
In the diagram, we can see that-
$\angle NMA+\angle AMO={{90}^{\circ }}$ (Complementary angles)
Given angle of depression for point A $={{60}^{\circ }}$ .
$\Rightarrow \angle NMA={{60}^{\circ }}$
So, $\begin{align}
& {{60}^{\circ }}+\angle AMO={{90}^{\circ }} \\
& \Rightarrow \angle AMO={{90}^{\circ }}-{{60}^{\circ }} \\
& \Rightarrow \angle AMO={{30}^{\circ }} \\
\end{align}$
We know that $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$
In \[\Delta OMA\] -
For $\angle AMO$ -
Perpendicular $=OA=xm$
Base $=OM=15m$ .
So, $\tan \angle AMO=\dfrac{OA}{OM}$ .
On putting $\angle AMO={{30}^{\circ }}$ , $OA=x.m$ and $OM=15m$ , we will get-
$\tan {{30}^{\circ }}=\dfrac{x}{15}$
We know $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ .
On putting $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ , we will get-
$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{x}{15}$ .
On multiplying both sides by 15, we will get-
$\Rightarrow \dfrac{15}{\sqrt{3}}=x$ .
On multiplying both numerator and denominator by $\sqrt{3}$ in LHS, we will get
$\begin{align}
& \Rightarrow \dfrac{15\sqrt{3}}{\sqrt{3}\times \sqrt{3}}=x \\
& \Rightarrow \dfrac{15\sqrt{3}}{3}=x \\
& \Rightarrow 5\sqrt{3}=x \\
& \Rightarrow x=5\sqrt{3}m \\
\end{align}$
Now let us find ‘y’
In the diagram, we can see that
$\angle NMB+\angle BMO={{90}^{\circ }}$ (Complementary angles)
Given angle of depression for point B$={{45}^{\circ }}$
$\Rightarrow \angle NMB={{45}^{\circ }}$
So, ${{45}^{\circ }}+\angle BMO={{90}^{\circ }}$
$\begin{align}
& \angle BMO={{90}^{\circ }}-{{45}^{\circ }} \\
& \angle BMO={{45}^{\circ }} \\
\end{align}$
Again, we will apply $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ for $\angle BMO$ in triangle OMB.
In $\Delta OMB$ -
Fro $\angle BMO$ -
Perpendicular $=OB=y.m$
Base $=OM=xm$ .
So, $\tan \angle BMO=\dfrac{OB}{OM}$ .
On putting $\angle BMO={{45}^{\circ }}$. $OB=y.m$ and $OM=15m$ , we will get-
$\tan {{45}^{\circ }}=\dfrac{ym}{15m}$ .
We know that $\tan {{45}^{\circ }}=1$ .
On putting $\tan {{45}^{\circ }}=1$ , we will get-
$1=\dfrac{y}{15}$ .
On multiplying both sides by 15, we will get-
$\begin{align}
& 15=y \\
& \Rightarrow y=15m \\
\end{align}$
Now, we have obtained $x=5\sqrt{3}m$ and $y=15m$ , So, putting $x=5\sqrt{3}m$ and $y=15m$ , we will get-
Distance between A and B \[=15-5\left( 1.732 \right)\]
$\begin{align}
& =15-8.660 \\
& =6.34 \\
\end{align}$
Hence, the required distance between points A and B is 6.34m.
Note: Students generally make mistakes in recognizing the angle of depression.
If from point ‘O’, we are watching point ‘A’, angle ‘∅’ is the angle of depression. But students get confused and take angle ‘α’ as an angle of depression.
Complete step by step answer:
According to question, we can draw the following diagram-
We have to find the distance between points ‘A’ and ‘B’. Let us assume the distance of point ‘A’ from the base of tower to be x.m and the distance of point B from the base of tower to be ‘y’m.
From the diagram, we can see that-
Distance between points ‘A’ and ‘B’ $=\left( y-x \right)m$ …………………………….. (1)
To find the distance between ‘A’ and ‘B’, we need to find ‘x’ and ‘y’.
Let us find ‘x’ first-
In the diagram, we can see that-
$\angle NMA+\angle AMO={{90}^{\circ }}$ (Complementary angles)
Given angle of depression for point A $={{60}^{\circ }}$ .
$\Rightarrow \angle NMA={{60}^{\circ }}$
So, $\begin{align}
& {{60}^{\circ }}+\angle AMO={{90}^{\circ }} \\
& \Rightarrow \angle AMO={{90}^{\circ }}-{{60}^{\circ }} \\
& \Rightarrow \angle AMO={{30}^{\circ }} \\
\end{align}$
We know that $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$
In \[\Delta OMA\] -
For $\angle AMO$ -
Perpendicular $=OA=xm$
Base $=OM=15m$ .
So, $\tan \angle AMO=\dfrac{OA}{OM}$ .
On putting $\angle AMO={{30}^{\circ }}$ , $OA=x.m$ and $OM=15m$ , we will get-
$\tan {{30}^{\circ }}=\dfrac{x}{15}$
We know $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ .
On putting $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ , we will get-
$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{x}{15}$ .
On multiplying both sides by 15, we will get-
$\Rightarrow \dfrac{15}{\sqrt{3}}=x$ .
On multiplying both numerator and denominator by $\sqrt{3}$ in LHS, we will get
$\begin{align}
& \Rightarrow \dfrac{15\sqrt{3}}{\sqrt{3}\times \sqrt{3}}=x \\
& \Rightarrow \dfrac{15\sqrt{3}}{3}=x \\
& \Rightarrow 5\sqrt{3}=x \\
& \Rightarrow x=5\sqrt{3}m \\
\end{align}$
Now let us find ‘y’
In the diagram, we can see that
$\angle NMB+\angle BMO={{90}^{\circ }}$ (Complementary angles)
Given angle of depression for point B$={{45}^{\circ }}$
$\Rightarrow \angle NMB={{45}^{\circ }}$
So, ${{45}^{\circ }}+\angle BMO={{90}^{\circ }}$
$\begin{align}
& \angle BMO={{90}^{\circ }}-{{45}^{\circ }} \\
& \angle BMO={{45}^{\circ }} \\
\end{align}$
Again, we will apply $\text{tan }\!\!\theta\!\!\text{ =}\dfrac{\text{perpendicular}}{\text{base}}$ for $\angle BMO$ in triangle OMB.
In $\Delta OMB$ -
Fro $\angle BMO$ -
Perpendicular $=OB=y.m$
Base $=OM=xm$ .
So, $\tan \angle BMO=\dfrac{OB}{OM}$ .
On putting $\angle BMO={{45}^{\circ }}$. $OB=y.m$ and $OM=15m$ , we will get-
$\tan {{45}^{\circ }}=\dfrac{ym}{15m}$ .
We know that $\tan {{45}^{\circ }}=1$ .
On putting $\tan {{45}^{\circ }}=1$ , we will get-
$1=\dfrac{y}{15}$ .
On multiplying both sides by 15, we will get-
$\begin{align}
& 15=y \\
& \Rightarrow y=15m \\
\end{align}$
Now, we have obtained $x=5\sqrt{3}m$ and $y=15m$ , So, putting $x=5\sqrt{3}m$ and $y=15m$ , we will get-
Distance between A and B \[=15-5\left( 1.732 \right)\]
$\begin{align}
& =15-8.660 \\
& =6.34 \\
\end{align}$
Hence, the required distance between points A and B is 6.34m.
Note: Students generally make mistakes in recognizing the angle of depression.
If from point ‘O’, we are watching point ‘A’, angle ‘∅’ is the angle of depression. But students get confused and take angle ‘α’ as an angle of depression.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)