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  • n Fig.10.13, angle ADC = 130° and chord BC = chord BE. Find angle CBE.

In Fig.10.13, \angleADC = 130° and chord BC = chord BE. Find \angleCBE.

 

Answers (1)

100°

Solution:

Consider \triangleBCO and \triangleBEO

\Rightarrow BC = BE as given.

\angleBCO = \angleBEO          (angles opposite to equal sides in a triangle are equal)

BO = BO                     (common sides)

 

\Rightarrow \triangleBCO \cong \triangleBEO     (SAS congruence)

\angleCBO = \angleOBE          (CPCT)           …(i)

Now we know that ABCD is a Cyclic quadrilateral,

So, \angle ADC + \angle ABC = 180°

\Rightarrow 130° + \angleABC = 180°

\Rightarrow \angle ABC = 180°-130° = 50°

\therefore \angle OBE = 50°

From (i)

\angle CBO =\angleOBE = 50°

\Rightarrow \angleCBE = \angle CBO +\angle OBE = 50° + 50° = 100°.

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