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Home » ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120degree Calculate: i) ∠BEC ii) ∠BED
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120degree Calculate: i) ∠BEC ii) ∠BED
Class 10 | Exercise 18C | ICSE | Maths | Selina | Tangents and Intersecting Chords
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120degree Calculate: i) ∠BEC ii) ∠BED

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(C) - 23

i) Join

    \[OC\text{ }and\text{ }OB\]

    \[AB\text{ }=\text{ }BC\text{ }=\text{ }CD\]

And

    \[\angle ABC\text{ }=\text{ }{{120}^{o}}~\]

[Given]

So,

    \[\angle BCD\text{ }=\angle ABC\text{ }=\text{ }{{120}^{o}}\]

    \[OB\text{ }and\text{ }OC\]

are the bisectors of 

    \[\angle ABC\text{ }and\angle BCD~\]

and respectively.

So,

    \[\angle OBC\text{ }=\angle BCO\text{ }=\text{ }{{60}^{o}}\]

In 

    \[\vartriangle BOC,\]

    \[\angle BOC\text{ }=\text{ }{{180}^{o}}-\text{ }(\angle OBC\text{ }+\angle BOC)\]

    \[\angle BOC\text{ }=\text{ }{{180}^{o}}-\text{ }({{60}^{o}}~+\text{ }{{60}^{o}})\]

    \[=\text{ }{{180}^{o}}-\text{ }{{120}^{o}}\]

So,

    \[\angle BOC\text{ }=\text{ }{{60}^{o}}\]

    \[Arc\text{ }BC\text{ }subtends~\angle BOC~\]

at the centre and 

    \[\angle BEC\]

 at the remaining part of the circle.

    \[\angle BEC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle BOC\]

    \[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{60}^{o}}~=\text{ }{{30}^{o}}\]

ii) In cyclic quadrilateral

    \[BCDE\]

We have

    \[\angle BED\text{ }+\angle BCD\text{ }=\text{ }{{180}^{o}}\]

    \[\angle BED\text{ }+\text{ }{{120}^{o}}~=\text{ }{{180}^{o}}\]

 

Thus,

    \[\angle BED\text{ }=\text{ }{{60}^{o}}\]

i

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