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  • In Fig. 10.21, O is the centre of the circle, BD = OD and CD \perp AB. Find \angleCAB.

In Fig. 10.21, O is the centre of the circle, BD = OD and CD \perp AB. Find \angleCAB.

Answers (1)

Given: In the figure BD = OD, CD\perp AB

In \triangleOBD,

BD = OD         (given)

OD = OB         (radius of the same circle)

OB = OD = BD

Hence the triangle is equilateral

\angleBOD = \angleOBD = \angleODB = 60°

Consider \triangleMBC and MBD

MB = MB                               (common)

\angleCMB = \angleBMD = 90°          (given)

CM = MD                               (perpendicular from the centre on the chord bisects the chord]

\triangleMBC =\triangleMBD                     (SAS rule)

\angleMBC =\angleMBD                    (CPCT)

\angleMBC = \angleOBD = 60°           (\becauseOBD = 60°)

Since AB is the diameter of the circle

\angleACB = 90°                           (angle is a semi-circle)

\angleCAB +\angleCBA + \angleACB = 180°       (By angle sum property of a triangle)

Putting the values,

\angleCAB + 60° + 90° = 180°

\angleCAB = 180° – (60° + 90°) = 30°

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