Two pillars of equal height stand on either side of a road-way which is 60 m wide. At a point in the road-way between the pillars, the elevation of the top of the pillars is $ {{60}^{\circ }} $ and $ {{30}^{\circ }} $ . The height of the pillar is
A. $ 15\sqrt{3}\text{ }m $
B. $ \dfrac{15}{\sqrt{3}}\text{ }m $
C. $ 15\text{ }m $
D. $ 20\text{ }m $
Answer
524.4k+ views
Hint: We first assume the height of the pillars. We then use trigonometric ratios for the given angles to find the relation between the base and the height of the pillars. We then form the equation and solve that to find the height of the pillars.
Complete step-by-step answer:
We first try to draw the scenario for the pillars.
The pillars are AC and BD. We assume the length of both of the pillars to be $ h $ .
The distance between the points A and B is 60 m. So, $ AB=60 $ .
At a point E in the road-way between the pillars, the elevation of the top of the pillars is $ {{60}^{\circ }} $ and $ {{30}^{\circ }} $ to the points C and D respectively. So, $ \angle CEA={{60}^{\circ }},\angle DEB={{30}^{\circ }} $ .
Now with respect to the angles $ \angle CEA={{60}^{\circ }},\angle DEB={{30}^{\circ }} $ and the equal sides of AC and BD, we find the length of the segments AE and BE.
We have
$ \begin{align}
& \dfrac{AE}{AC}=\cot \left( \angle CEA \right)=\cot {{60}^{\circ }} \\
& \dfrac{BE}{BD}=\cot \left( \angle DEB \right)=\cot {{30}^{\circ }} \\
\end{align} $
We place the values to get
$ \begin{align}
& \dfrac{AE}{h}=\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}} \\
& \dfrac{BE}{h}=\cot {{30}^{\circ }}=\sqrt{3} \\
\end{align} $
On simplification we get $ AE=\dfrac{h}{\sqrt{3}},BE=h\sqrt{3} $ .
We know $ AE+BE=AB=60 $ . Putting the values, we get
$ \begin{align}
& \dfrac{h}{\sqrt{3}}+h\sqrt{3}=60 \\
& \Rightarrow \dfrac{4h}{\sqrt{3}}=60 \\
& \Rightarrow h=\dfrac{60\times \sqrt{3}}{4}=15\sqrt{3} \\
\end{align} $
The length of the pillars is $ 15\sqrt{3} $ . The correct option is A.
So, the correct answer is “Option A”.
Note: In the relation we will take that particular trigonometric ratio which gives the unknowns as the numerator form. That particular ratio gives a direct answer for the unknown. Also, we could use the ratios as the triangles were right angle triangles.
Complete step-by-step answer:
We first try to draw the scenario for the pillars.
The pillars are AC and BD. We assume the length of both of the pillars to be $ h $ .
The distance between the points A and B is 60 m. So, $ AB=60 $ .
At a point E in the road-way between the pillars, the elevation of the top of the pillars is $ {{60}^{\circ }} $ and $ {{30}^{\circ }} $ to the points C and D respectively. So, $ \angle CEA={{60}^{\circ }},\angle DEB={{30}^{\circ }} $ .
Now with respect to the angles $ \angle CEA={{60}^{\circ }},\angle DEB={{30}^{\circ }} $ and the equal sides of AC and BD, we find the length of the segments AE and BE.
We have
$ \begin{align}
& \dfrac{AE}{AC}=\cot \left( \angle CEA \right)=\cot {{60}^{\circ }} \\
& \dfrac{BE}{BD}=\cot \left( \angle DEB \right)=\cot {{30}^{\circ }} \\
\end{align} $
We place the values to get
$ \begin{align}
& \dfrac{AE}{h}=\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}} \\
& \dfrac{BE}{h}=\cot {{30}^{\circ }}=\sqrt{3} \\
\end{align} $
On simplification we get $ AE=\dfrac{h}{\sqrt{3}},BE=h\sqrt{3} $ .
We know $ AE+BE=AB=60 $ . Putting the values, we get
$ \begin{align}
& \dfrac{h}{\sqrt{3}}+h\sqrt{3}=60 \\
& \Rightarrow \dfrac{4h}{\sqrt{3}}=60 \\
& \Rightarrow h=\dfrac{60\times \sqrt{3}}{4}=15\sqrt{3} \\
\end{align} $
The length of the pillars is $ 15\sqrt{3} $ . The correct option is A.
So, the correct answer is “Option A”.
Note: In the relation we will take that particular trigonometric ratio which gives the unknowns as the numerator form. That particular ratio gives a direct answer for the unknown. Also, we could use the ratios as the triangles were right angle triangles.
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

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

