Question

In the given figure , ∠ ABC = 130 ° and chord BC = chord BE. Find ∠CBE.​

Answer 1

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$\large\bf \red{Answer}$

$\small \tt{Let \: us \: consider \: , the \: points \: A , B , C \: and \: D \: from } \\ \small \tt{ Cyclic \: quadrilateral .}$

$\small \tt{ \therefore \angle \: AD + \angle \: OBC = 180 \degree[opposite \: angle \: of \: cyclic \: quadrilateral ].}$

$\small \tt{{ \implies \: 130 \degree + \angle \:OBC = 180 \degree }}$

$\small \tt{{ \implies \angle \: OBC = 180 \degree - 130 \degree}}$

$\small \tt{{ \implies \angle \: OBC = 50 \degree}}$

$\tt \small{In \: \triangle \: BOC \: and \: \triangle \: BOE,}$

$\tt \small \: BC = BE \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [Given]$

$\tt \small{OC = OE} \: \: \: \: \: \: \: \: \: [Radius \: of \: a \: same \: circle]$

$\tt \small{and \: \: OB= OB \: \: \: \: \: \: \: \: \: \: [Common]} \\\tt \small{\therefore\triangle \: BOC= \triangle \: BOE \: \: \: [By \: SSS \: congruency] }$

$\tt \small{ \implies\angle \: OBC= \angle \: OBE \: \: \: \: \: \: [ By \: C.P.C.T ]}$

$\tt \small \: { Now, \angle \: CBE= \angle \: CBO+ \angle \: EBO}$

$\tt \small{ = 50 \degree + 50 \degree = 100 \degree}$

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