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Question

ABC is an equilateral triangle and O is its circumcentre. If the side of triangle is 6 cm, then the $\angle BOC$ is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$120^\circ$

Problem Analysis:

  • We are given an equilateral triangle ABC.
  • O is the circumcentre of this triangle.
  • The side length is 6 cm.
  • We need to find the measure of the angle $\angle BOC$.

Key Geometric Principle:

The angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc at any point on the circumference. In this case, arc BC subtends $\angle BOC$ at the centre O and $\angle BAC$ at the circumference (point A).

Applying the Principle:

In an equilateral triangle, each interior angle measures $60^\circ$. Therefore:

$ \angle BAC = 60^\circ $

Using the angle theorem:

$ \angle BOC = 2 \times \angle BAC $

Calculation:

  1. Substitute the value of $\angle BAC$: $ \angle BOC = 2 \times 60^\circ $
  2. Calculate the final angle: $ \angle BOC = 120^\circ $

Note: The side length of 6 cm is not required to determine the angle $\angle BOC$.

Conclusion:

The angle $\angle BOC$ is $120^\circ$. This corresponds to Option D.

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Similar Questions

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  3. PQ is a diameter of circle whose centre is O. If a point R lies on a circle and $\angle RPO$ is $42^\circ$, then find $\angle RQP$.
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Important Questions from Geometry

  1. ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?

  2. If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:

  3. If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.  

  4. If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?

  5. D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.

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