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  • In Fig. 10.16, ∠OAB = 30º and ∠OCB = 57º. Find ∠BOC and ∠AOC.

In Fig. 10.16, \angleOAB = 30º and \angleOCB = 57º. Find \angleBOC and \angleAOC.

Answers (1)

Given: \angleOAB = 30°, \angleOCB = 57°C

In \triangleOAB,

AO = BO (radius of the same circle)

\angleOAB = \angleOBA = 30° (Angles opposite to equal sides are equal)

In \triangleAOB, the sum of all angles is 180°.

\Rightarrow \angleOAB + \angleOBA + \angleAOB = 180°.

\Rightarrow 30° + 30° + \angleAOB = 180°

\Rightarrow \angleAOB = 180° – 30° – 30°

\Rightarrow \angleAOB = 120°                    … (i)

Now, in \triangleOBC,

OC = OB (radius of the same circle)

\angleOBC = \angleOCB = 57° (Angles Opposite to equal sides are equal).

In \triangleOBC, the sum of all angles is 180°.

\angleOBC + \angleOCB + \angleBOC = 180°

\Rightarrow 57° + 57° + \angleBOC = 180°

\angleBOC = 180° – 57° – 57°

\Rightarrow \angleBOC = 66° … (ii)

Now, from equation (i) we have

\Rightarrow \angleAOB = 120°.

\Rightarrow \angleAOC + \angleCOB = 120°

\Rightarrow \angleAOC + 66° = 120° (from ii)

\Rightarrow \angleAOC = 120° – 66°

\Rightarrow\angleAOC = 54° and \angle BOC = 66°.

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