To find ∠ADC when A, B, and C are points on a circle with center O, and ∠BOC = 30° and ∠AOB = 60°, and D is a point on the circle other than the arc ABC, we use the properties of angles in a circle.
∠BOC and ∠AOB are angles at the center of the circle. The angle at the center is twice the angle at the circumference on the same arc. Therefore, ∠BAC, which stands on arc BC, is half of ∠BOC, so ∠BAC = 15°. Similarly, ∠ABC, standing on arc AC, is half of ∠AOB, so ∠ABC = 30°.
In triangle ABC, the sum of angles is 180°. With ∠BAC = 15° and ∠ABC = 30°, ∠ACB = 180° – 15° – 30° = 135°.
Now, consider ∠ADC. D and B are points on the circle in the segment not containing A. The angle in a segment is constant, so ∠ADC = ∠ACB = 135°.

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## Circle Geometry and Angle Relationships

In the realm of circle geometry, the relationships between angles formed by points on a circle and the circle’s center are fundamental. In this scenario, we have a circle with center O and points A, B, C, and D on its circumference. The given angles are ∠BOC = 30° and ∠AOB = 60°. Our objective is to determine the measure of ∠ADC, where D is a point on the circle other than the arc ABC. This problem is a classic example of applying the principles of circle geometry to find unknown angles.

### Understanding Central and Circumferential Angles

Central angles in a circle, like ∠BOC and ∠AOB, have a unique property: they are twice the size of the angles at the circumference that stand on the same arc. This principle is key to solving our problem. ∠BOC and ∠AOB are central angles, and their corresponding circumferential angles will be half of their measures. This relationship allows us to calculate the angles at points A, B, and C on the circumference of the circle.

#### Calculating Angles at the Circumference

Using the property that the angle at the center is twice the angle at the circumference, we find that ∠BAC, which subtends the same arc as ∠BOC, is half of 30°, thus ∠BAC = 15°. Similarly, ∠ABC, which subtends the same arc as ∠AOB, is half of 60°, making ∠ABC = 30°. These calculations are crucial for understanding the angular relationships in the circle and for determining the measure of ∠ADC.

Determining the Angle in Triangle ABC
In triangle ABC, the sum of the internal angles is always 180°. We already know two angles: ∠BAC = 15° and ∠ABC = 30°. Therefore, the third angle, ∠ACB, can be calculated as 180° – 15° – 30° = 135°. This angle is significant because it will help us in determining the measure of ∠ADC.

##### Applying the Angle in the Same Segment Rule

The angle in the same segment rule states that angles in the same segment of a circle are equal. Since points D and B lie in the same segment of the circle, separated by the chord AC, the angle ∠ADC will be equal to ∠ACB. This is because both angles are subtended by the same arc AC.

###### Finding the Measure of ∠ADC

By applying the principles of circle geometry, we conclude that ∠ADC, which is in the same segment as ∠ACB and subtended by the same arc, is equal to ∠ACB. Therefore, ∠ADC = 135°. This solution not only demonstrates the application of geometric principles but also highlights the elegance and consistency of mathematical relationships in circle geometry. Understanding these concepts allows for a deeper appreciation of the symmetry and patterns inherent in geometric shapes.

Discuss this question in detail or visit to Class 9 Maths Chapter 9 for all questions.
Questions of 9th Maths Exercise 9.3 in Detail

 In Figure, A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC. 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. In Figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR. In Figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC. In Figure, 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. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD. 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. Prove that ∠ACP = ∠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 triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD. Prove that a cyclic parallelogram is a rectangle.

Last Edited: January 6, 2024