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If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC. - Mathematics

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प्रश्न

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC = 120°, prove that BC + BD = BO, i.e., BO = 2BC.

योग

उत्तर


To prove: BO = 2BC

Given, ∠DBC = 120°

Join OC, OD and BO.

Since, BC and BD are tangents.

∴ OC ⊥ BC and OD ⊥ BD

We know, OB is the angle bisector of ∠DBC.

∴ ∠OBC = ∠DBO = 60°

In right-angled ∆OBC,

cos 60° = `("BC")/("OB")`

⇒ `1/2 = ("BC")/("OB")`

⇒ OB = 2 BC

Also, BC = BD   ...[Tangent drawn from external point to circle are equal]

OB = BC + BC

⇒ OB = BC + BD

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Circles - Exercise 9.3 [पृष्ठ १०७]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 10
अध्याय 9 Circles
Exercise 9.3 | Q 3 | पृष्ठ १०७
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