An isosceles triangle of vertical angle 2a is inscribed in a circle of radius r.Show that the area of triangle is maximum when a=(pi)/6.

given: ABC is an isosceles triangle such that AB=AC. the vertical angle ∠BAC =2a. and triangle is inscribed in the circle with center O and radius r.

TPT: area(ΔABC) is maximum when a=

since the triangle is an isosceles triangle . the circum center of the circle will lie on the perpendicular from A to BC.

let it be AM.

∠BOC = 2*2a= 4a [angle subtended at the center is twice the angle subtended at the circle]

∠BOM = 2a [since ΔOMB and OMC are congruent triangles]

OA=OB=OC =r [radius of the circle]

now in the triangle OMB,

BC= 2BM [perpendicular from the center bisects the chord]

BC=2r sin2a..........(1)

height of the triangle ΔABC = AM = AO+OM

AM= r +rcos2a ..............(2)

area of the ΔABC A =

..............(3)

since r (the radius of the circle ) is fixed.

differentiating (3) wrt a :

........(4)

differentiating again wrt a:

.........(5)

for maximum value of area equating dA/da = 0

if 2a = π it will not form the triangle.

for a=π/6, is negative therefore for a=π/6 area is maximum.

thus for the maximum area triangle must be an equilateral triangle.

hope this helps you.

**
**