In Figure, A,B and C are three points on a circle with centre O

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on January 6, 2024, 11:00 AM

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°.

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,

Let’s discuss in detail

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.