हिंदी

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°. Calculate : ∠ADC Also, show that the ΔAOD is an equilateral triangle. - Mathematics

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प्रश्न

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°. 

Calculate : ∠ADC 

Also, show that the ΔAOD is an equilateral triangle.

योग

उत्तर


∠ABD + ∠DBC = 30° + 30° = 60° 

`=>` ∠ABC = 60°

In cyclic quadrilateral ABCD,

∠ADC + ∠ABC = 180°

(Pair of opposite angles in a cyclic quadrilateral are supplementary)

`=>` ∠ADC = 180° – 60° = 120°

In ∆AOD, OA = OD (Radii of the same circle)

∠AOD = ∠DAO Or ∠DAB = 60° [Proved in (i)]

∠AOD = 60°

`=>` ∠ADO = ∠AOD = ∠DAO = 60°

∴ ∆AOD is an equilateral triangle.

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अध्याय 17: Circles - Exercise 17 (A) [पृष्ठ २६१]

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सेलिना Mathematics [English] Class 10 ICSE
अध्याय 17 Circles
Exercise 17 (A) | Q 38.4 | पृष्ठ २६१

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