An isosceles triangle of vertical angle 2\theta is inscribed in a circle of radius a. Show the area of the triangle is maximum when \theta = \frac{\pi }{6}\cdot

 

 

 

 
 
 
 
 

Answers (1)

consider the figure shown.

Let OD= x
Here CD= \sqrt{a^{2}-x^{2}}
Now arc \left ( ABC \right )= A= \frac{1}{2}\times 2\sqrt{a^{2}-x^{2}} \times \left ( a+x \right )
Let s= A^{2}= \left ( a^{2}-x^{2} \right )-\left ( a+x \right )^{2}= \left ( a-x \right )\left ( a+x \right )^{3}
\Rightarrow \frac{ds}{dx}= \left ( a-x \right )\times 3\left ( a+x \right )^{2}+\left ( a+x \right )^{3}\left ( -1 \right )
\Rightarrow \frac{ds}{dx}= 2\left ( a-2x \right )\left ( a+x \right )^{2}
\Rightarrow \frac{d^{2}s}{dx^{2}}= 4\left ( a-2x \right )\left ( a+x \right )-4\left ( a+x \right )^{2}

for \frac{ds}{dx}= 2\left ( a-2x \right )\left ( a+x \right )^{2}= 0\; \; \; \therefore x= \frac{9}{2}\; \left [ \because x\neq -a \: \: As \frac{d^{2}s}{dx^{2}} \right ]_{at} x= \frac{a}{2}= -9a^{2}< 0
\therefore s is maximum at x= \frac{a}{2}
\therefore A is also maximum
Now in \bigtriangleup CDA= \tan \theta = \frac{CD}{AD}= \frac{\sqrt{a^{2}-\frac{a^{2}}{4}}}{a+\frac{a}{2}}= \frac{a\times \frac{\sqrt{3}}{2}}{a\times \frac{\sqrt{3}}{2}}= \frac{1}{\sqrt{3}}
\therefore \theta = \frac{\pi }{6}

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