
In ∆OAB,
AB = OA = OB = radius
∠∆OAB is an equilateral triangle.
Therefore, each interior angle of this triangle will be of 60°.
∠AOB = 60°
∠ACB=\(\frac{1}{2}\)∠AOB=\(\frac{1}{2}\)(60°)=30°
In cyclic quadrilateral ACBD,
∠ACB + ∠ADB = 180° (Opposite angle in cyclic quadrilateral)
∠ADB = 180° − 30° = 150°
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.

In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.

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 Fig. 9.27). Prove that ∠ACP = ∠ QCD

Section | Number of girls per thousand boys |
|---|---|
Scheduled Caste (SC) | 940 |
Scheduled Tribe (ST) | 970 |
Non-SC/ST | 920 |
Backward districts | 950 |
Non-backward districts | 920 |
Rural | 930 |
Urban | 910 |
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
