Advertisements
Advertisements
प्रश्न
In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
Advertisements
उत्तर
In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180° ...(Angle sum property of a triangle)
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠BAC = 180° − 100º
⇒ ∠BAC = 80°
∠BDC = ∠BAC = 80° ...(Angles in the same segment of a circle are equal)
APPEARS IN
संबंधित प्रश्न
Fill in the blank:
A circle divides the plane, on which it lies, in ............ parts.
Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.
Given an arc of a circle, show how to complete the circle.
In the below fig. O is the centre of the circle. Find ∠BAC.
If O is the centre of the circle, find the value of x in the following figure
In the given figure, it is given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.
Prove that the angle in a segment greater than a semi-circle is less than a right angle.
In the given figure, two circles intersect at A and B. The centre of the smaller circle is Oand it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.
In the given figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =
A circle has radius `sqrt(2)` cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45°.
