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In figure, `angleADC=130^(@)" and chord BC=chord BE. Find "angleCBE`. |
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Answer» We have, `angleADC=130^(@)` and chord BC=chord BE. Suppose, we consider the points A, B, C and D form a cyclic quadrilateral. Since, the sum of opposite angles of a cyclic quadrilateral ADCB is ` 180^(@)`. `:. angleADC+angleOBC=180^(@)` `rArr 130^(@)+angleOBC=180^(@)` `rArr angleOBC=180^(@)-130^(@)=50^(@)` `"In" DeltaBOC and DeltaBOE`, BC=BE [given equal chord] OC=OE [both are the radius of a circle] and OB=OB [common side] `:. DeltaBOC cong DeltaBOE` [by SSS songruence rule] `rArr angle OBC=angleOBE=50^(@)` [by CPCT] Now, `angleCBE=angleCBO+angleEBO` `=50^(@)+50^(@)=100^(@)` |
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