Circle geometry - 24th Nov

Question

!!ABC!! is an isosceles triangle with !!BC=1!! inscribed in a circle of unit radius. What is !!PQ?!!

!!a)" "2(2-sqrt3)!!

!!b)" "(2-sqrt3)!!

!!c)" "2(2+sqrt3)!!

!!d)" "!!None

Solution

In the given figure:

!!DeltaABC!! is an isosceles traingle and !!o!! is the center of the circle.

Now through the property of the circle !!angleBOC = 2angleA!!

We can see that since !!OB = OC = 1!!, hence !!DeltaBOC!! is also an isosceles triangle. This will give !!angleOBC = 90 - angleA!!

In !!DeltaABC!!, !!angleB = angleC = 90 - angle(A/2)!!

So !!angleOBA = angleB - angleOBC!!

!!=>angleOBA = 90 - angle(A/2) - (90 - angleA) = angle(A/2)!!........(i)

Also in !!DeltaAPQ!!, !!angleAPQ = 90 - angle(A/2)!!

Using straight line concept !!APB!!, we get !!angleOPB = 180 - angleAPQ = 180 - 90 + angle(A/2) = 90 + angle(A/2)!!.....(ii)

In !!DeltaBOC!!, Using triangle property

!!(BC)/(Sin2A) = (OC)/(SinOBC)!!

!!=> 1/Sin2A = 1/Sin(90 - angleA)!!

!!=> SinA = 1/2!!

!!=> A = 30!!

Using this in (i) and (ii), we get

!!angleOPB = 90 + 15!! and !!angleOBA = angle(A/2) = 15!!

Now Using triangle property in !!DeltaOPB!!, we have

!!(OP)/(SinOBA) = (OB)/(SinOPB)!!

!!=> (OP)/Sin15 = 1/cos15!!

!!=> OP = tan15!!

Hence, !!PQ = 2OP = 2tan15 = 2tan(45-30) = 2(2-sqrt3)!!