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Question
In the figure, ▢ABCD is a cyclic quadrilateral. If m(arc ABC) = 230°, then find ∠ABC, ∠CDA, ∠CBE.
Solution
m(arc ABC) = 230° .....(i) [Given]
∴ m(arc ADC) + m(arc ABC) = 360° .......[Degree measure of a circle is 360°]
∴ m(arc ADC) = 360° – m(arc ABC)
∴ m(arc ADC) = 360° – 230° .......[From (i)]
∴ m(arc ADC) = 130°
∠ABC = `1/2` m (arc ADC) ......[Inscribed angle theorem]
= `1/2 xx 130^circ`
= 65°
Now, ∠CDA = `1/2` m (arc ABC) ......[Inscribed angle theorem]
∴ ∠CDA = `1/2 xx 230^circ`
∴ ∠CDA = 115° ......(ii)
∠CBE = ∠CDA ......(iiii) [The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle]
∴ ∠CBE = 115° .....[From (ii) and (iii)]
∴ ∠ABC = 65°, ∠CDA = 115°, ∠CBE = 115°.
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