Question

If O is the center of the circle, find the value of x in each of the following figures. pls do step by step

Answer 1

In 1st figure;

In ∆AOB,∆B

OA = OB = radius of circle

=>OAB = OBA = x°

AOB = 180 - 120 = 60°

Therefore,

AOB + OAB + OBA = 180

2x = 180 - 60 = 120

x = 120 ÷ 2 = 60

In 2nd figure;

In ∆AOC and ∆BOD,

OB = OA [Both are radius of circle]

BOD = AOC. [Vertically opposite angles]

OD = OC [Both are radius if circle]

So, ∆AOC is congruent to ∆BOD [SAS]

So, x = 50° [Alternate interior angle]

In 3rd figure;

COE = CBE = 50°. [Angles in the same segment]

ACB = 90° [Angle in a semi-circle]

so, BCD = 180 ° - 90° = 90°

Now, In ∆BCD,

90° + 50° + x = 180°

x = 180 - 140° = 40°

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