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  • In Fig. 10.20, O is the centre of the circle, angle BCO = 30 degree. Find x and y.

In Fig. 10.20, O is the centre of the circle, \angleBCO = 30°. Find x and y.

Answers (1)

x = 30° and y = 15°

Solution:

Given: O is the centre of the circle

\angleBCO = 30°

Join OB and AC.

In \triangleBOC,

CO = BO (Radius of the same circle)

\angleOBC =\angleOCB = 30°  (angles opposite to equal sides in a triangle are equal)

In \triangleOBC,

\angleOBC + \angleOCB + \angleBOC = 180°  (angle sum property of a triangle)

\angleBOC = 180° – (30° + 30°) = 120°

\angleBOC = 2\angleBAC (Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)

\angle BAC = \frac{120^{\circ}}{2}=60^{\circ}

Consider \triangleAEB, \triangleAEC

AE = AE (common)

\angleAEB =\angleAEC = 90°

BE = EC (given)

\triangleAEB \cong \triangleAEC  (SAS congruence)

\angleBAE =\angleCAE  (CPCT)

\angleBAE + \angleCAE = \angleBAC = 60°

2\angleBAE = 60°

\angleBAE = 30° = x

Now, OD is perpendicular to AE

and, CE is perpendicular to AE

Two lines perpendicular to the same line are parallel to each other

So, OD || CE and OC is transversal

\angleDOC = \angleECO = 30°  (alternate interior angles)

Now, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle.

Consider the arc CD,

\angleDOC = 2\angleDBC = 2y

30° = y

y = 15°

Hence, x = 30° and y = 15°

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infoexpert24

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