In a circle of radius #'r'#, chords of lengths #a# and #b# cms, subtend angles #q# and #3q# respectively at the centre, then #r = asqrt(a/(2a-b))# cm. True or False?

1 Answer
P dilip_k
Dec 13, 2017

The given relation is false

Explanation:

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We know that perpendicular dropped from the centre of the circle to a chord bisects the chord and the angle subtended by the chord at the centre of the circle.

So here we have

#/_AOM=1/2/_AOB and AM=1/2AB#
If #/_AOB=x , OA =r and AB=d# then #sin(x/2)=(AM)/(OA)=(d/2)/r=d/(2r)#
By the problem

When #d=a and x=q#

We get

#sin(q/2)=a/(2r)......(1)#

Again when #d=b and x=3q#

We get

#sin((3q)/2)=b/(2r)......(2)#

Dividing (2) by (1) we get

#sin((3q)/2)/sin(q/2)=b/a#

#=>(3sin(q/2)-4sin^3(q/2))/sin(q/2)=b/a#

#=>3-4sin^2(q/2))=b/a#

#=>3-4(a/(2r))^2=b/a#

#=>3-a^2/r^2=b/a#

#=>a^2/r^2=3-b/a#

#=>a^2/r^2=(3a-b)/a#

#=>r^2=a^3/(3a-b)#

#=>r=asqrt(a/(3a-b))#

Hence the given relation is false

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