Answer :

Let BC be chord, which is equal to the radius. Join OB and OC.

Given:

BC = OB = OC

\(\therefore, \) \(\triangle{OBC}\) is an equilateral traingle.

\(\therefore \) \(\angle{BOC}\) = \(60^\circ\)

But, we know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any part of the circle.

\(\therefore \) \(\angle{BAC}\) = (\(\frac{1}{2} \) ) \(\angle{BOC}\)

\(\Rightarrow \) \(\angle{BAC}\) = (\(\frac{1}{2} \) ) \(60^\circ\) = \(30^\circ\)

Also, here, ABMC is a cyclic quadrilateral.

\(\therefore \) \(\angle{BAC}\) + \(\angle{BMC}\) = \(180^\circ\)

(Since, in a cyclic quadrilateral the sum of opposite angles is \(180^\circ\)).

\(\therefore \) \(\angle{BMC}\) = \(180^\circ\) - \(30^\circ\) = \(150^\circ\)

Hence, the angle subtended by
the chord at a point on the minor arc and also at a point on the major arc is \(150^\circ\).

- In figure A, B and C are three points on a circle with centre O such that \(\angle{BOC}\) = \(30^\circ\) and \(\angle{AOB}\) = \(60^\circ\). If D is a point on the circle other than the arc ABC, find \(\angle{ADC}\).Let PQ = QR = PR = X
- In figure, \(\angle{PQR}\) = \(100^\circ\) ,where P, Q and R are points on a circle with centre O. Find \(\angle{OPR}\).
- In figure, \(\angle{ABC}\) = \(69^\circ\), \(\angle{ACB}\) = \(31^\circ\). Find \(\angle{BDC}\).
- In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that \(\angle{BEC}\) = \(130^\circ\) and \(\angle{ECD}\) = \(20^\circ\). Find \(\angle{BAC}\).
- ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \(\angle{DBC}\) = \(70^\circ\), \(\angle{BAC}\) = \(30^\circ\), Find \(\angle{BCD}\). Further, if AB = BC, find \(\angle{EDC}\).
- If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
- If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
- Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that \(\angle{ACP}\) = \(\angle{QCD}\).
- If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
- ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that \(\angle{CAD}\) = \(\angle{CBD}\).
- Prove that a cyclic parallelogram is a rectangle.

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- NCERT solutions for class 9 maths chapter 2 Polynomials
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