# An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when . Δ ABC is an isosceles triangle such that AB = AC.

The vertical angle BAC = 2θ

Triangle is inscribed in the circle with center O and radius a.

Draw AM perpendicular to BC.

Since, Δ ABC is an isosceles triangle, the circumcenter of the circle will lie on the perpendicular from A to BC.

Let O be the circumcenter.

BOC = 2×2θ = 4θ (Using central angle theorem)

COM = 2θ (Since, Δ OMB and Δ OMC are congruent triangles)

OA = OB = OC = a (radius of the cicle)

In Δ OMC,

CM = asin2θ

OM = acos2θ

BC = 2CM (Perpendicular from the center bisects the chord)

BC = 2asin2θ

Height of Δ ABC = AM = AO + OM

AM = a + acos2θ Differentiation this equation with respect to θ      Maxima or minima exists when: Therefore,    Therefore,     To check whether which point has a maxima, we have to check the double differentiate.

Therefore, at θ = :  Both the sin values are positive. So the entire expression is negative. Hence there is a maxima at this point.

θ = will not form a triangle. Hence it is discarded.

There fore the maxima exits at: Rate this question :

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