Exams › JEE Main › Maths
A triangle ABC is inscribed in a circle with centre O, and the angle at A is 60 degrees. From O a perpendicular OD is dropped onto the chord BC. Find the measure of angle BOD.
- 30 deg
- 45 deg
- 60 deg
- 90 deg
Correct answer: 60 deg
Solution
The central angle BOC = 2 * angle A = 2 * 60 = 120 degrees (angle at centre is twice the angle at the circumference on the same arc). Since OD is perpendicular to chord BC, it bisects angle BOC, so angle BOD = 120/2 = 60 degrees.
Related JEE Main Maths questions
- In a circle, the chord ED is parallel to the diameter AC, and angle CBE = 50 degrees (B is a point on the circle). Find angle CED.
- From an external point B, two lines BA and BC are drawn tangent to a circle, touching it at A and C respectively. E is a point on the major arc AC such that the inscribed angle AEC = 110 deg. Find the angle ABC.
- Two chords TY and OP of a circle intersect at point K such that TK = 2, KY = 16, and KP = 2*(KO). Find the length OP.
- Let A be the set {(n, 2n): n ∈ N} and let B be the set {(2n, 3n): n ∈ N}. What is the intersection A ∩ B?
- Let set A contain 3 elements and set B contain 6 elements. Then the cardinality of their union must satisfy
- Given the sets A = {1, 2, 5} and B = {3, 4, 5, 9}, what is A ∩ B?
⚔️ Practice JEE Main Maths free + battle 1v1 →