JEE Mains · Maths · STD 11 - Trigonometrical equations
A pole stands vertically inside a triangular park \(ABC\). Let the angle of elevation of the top of the pole from each corner of the park be \(\frac{\pi}{3}\). If the radius of the circumcircle ot \(\Delta ABC\) is \(2 ,\) then the height of the pole is equal to :
- A \(\frac{2 \sqrt{3}}{3}\)
- B \(2 \sqrt{3}\)
- C \(\sqrt{3}\)
- D \(\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Let \(P D=h, R=2\) As angle of elevation of top of pole from \(A , B , C\) are equal \(So\) \(D\) must be circumcentre of \(\triangle ABC\) \(\tan \left(\frac{\pi}{3}\right)=\frac{ PD }{ R }=\frac{ h }{ R }\) \(h=R \tan \left(\frac{\pi}{3}\right)=2 \sqrt{3}\)
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