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)