AB is the diameter and AC is a chord of a circle with centre O such that angle BAC- 30 - The tangent to the circle at C intersect AB product at- D- Show that BC-BD -

Let O be the centre of the circle.By tangent chord theorem #8736;CAB = #8736;BCD=30 #176; #160;Also #8736;OCD=90 #176;(Angle between tangent and ra ...

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Standard XII

Mathematics

Direct Common Tangent

AB is the dia...

Question

AB is the diameter and AC is a chord of a circle with centre $O$ such that angle $B A C = 30^{\circ}$ The tangent to the circle at $C$ intersect $A B$ product at. $D$ . Show that $B C = B D$ ?

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Solution

Let O be the centre of the circle.

By tangent chord theorem∠CAB =∠BCD=30°

Also ∠OCD=90°(Angle between tangent and radius is 90 degree)

∴∠BCO $= 90 ° - 30 ° = 60 °$ . Also ∠CBO=60° because,Angle opposite to equal sides are equal.

And hence ΔOBC must be a Equilateral triangle ∴∠COB=60°.

Also in Right Triangle OCD, ∠CDO $= 90 ° - 60 ° = 30 °$

So that BC=BD

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AB is the diameter and AC is a chord of a circle with centre O such that angle BAC=30∘ The tangent to the circle at C intersect AB product at.D. Show that BC=BD ?

AB is the diameter and AC is a chord of a circle with centre $O$ such that angle $B A C = 30^{\circ}$ The tangent to the circle at $C$ intersect $A B$ product at. $D$ . Show that $B C = B D$ ?