A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Here, the chord AB is equal to the radius of the circle. In the above diagram, OA and OB are the two radii of the circle.
Now, consider the ΔOAB.
AB = OA = OB = radius of the circle.
So, it can be said that ΔOAB has all equal sides and thus, it is an equilateral triangle.
∴ AOC = 60°
And, ACB = \(\frac{1}{2}\) AOB
ACB = \(\frac{1}{2}\) × 60° = 30°
Now, since ACBD is a cyclic quadrilateral,
ADB + ACB = 180° (Since they are the opposite angles of a cyclic quadrilateral)
ADB = 180°-30° = 150°
So, the angle subtended by the chord at a point on the minor arc and also at a point on the major arc are 150° and 30° respectively.
-
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
3 years ago
-
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
3 years ago
-
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.
3 years ago
-
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
3 years ago
-
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE, = AD.
3 years ago
- 321 Forums
- 27.3 K Topics
- 53.8 K Posts
- 0 Online
- 12.4 K Members