Question

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Answer 1

Answer:

a. Given: ABCD is a cyclic, angle A is 100 degrees

By the property of cyclic, the sum of alternate angles of a cyclic is 180

angle A + angle C = 180

100+ C = 180

C = 80 degree

b. Given: angle PRQ = 62 degree

Since the theorem states that and subtended at the centre of the circle by an arc is double the angle subtended by the same arc at any other point of the circle.

Therefore, angle POQ = 2 PRQ

POQ = 2 x 62 => 124 degree

c. Given angle DBC = 56 degree

Join CD with a line segment

Since angles subtended at two points by a line segment are same

Therefore, angle DBC = angle CAD

angle CAD = 56 degree

d. M is the midpoint of AB

Given: OA= 5 cm OM= 3 cm

Since, OA and OB are the radii of the circle

Therefore, OA = OB = 5 cm

In triangle OMA,

Using the Pythagoras theorem

AO^2 = OM^2 + AM^2

25 = 9 + AM^2

16 = AM^2

AM = √16

AM = 4 cm

Since M is the midpoint of AB

Therefore, it divides AB in the ratio 1:1

So, AM = MB = 4 cm

Thus, AB = 4 + 4 = 8 cm