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Question
An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. Show that the area of triangle is maximum when θ = `pi/6`
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Solution
Let ABC be an isosceles triangle inscribed in the circle with radius a such that AB = AC.
AD = AO + OD = a + a cos2θ and BC = 2BD = 2a sin2θ (see fig. 16.4)
Therefore, area of the triangle ABC
i.e. ∆ = `1/2` BC . AD
= `1/2 2"a" sin2theta * ("a" + "a" cos2theta)`
= a2sin2θ (1 + cos2θ)
⇒ ∆ = `"a"^2sin2theta + 1/2 "a"^2 sin4theta`
Therefore, `("d"∆)/("d"theta)` = 2a2cos2θ + 2a2cos4θ
= 2a2(cos2θ + cos4θ)
`("d"∆)/("d"theta)` = cos2θ = –cos4θ = cos (π – 4θ)
Therefore, 2θ = π – 4θ
⇒ θ = `pi/6`
`("d"^2∆)/("d"theta)` = 2a2 (–2sin2θ – 4sin4θ) < 0 `("at" theta = pi/6)`.
Therefore, Area of triangle is maximum when θ = `pi/6`.
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