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Question: A triangle has given base and a given perimeter. Then the vertical angle is greatest when the triang...

A triangle has given base and a given perimeter. Then the vertical angle is greatest when the triangle.

A

Is equilateral

B

Is isosceles

C

Has different side

D

None of these

Answer

Is isosceles

Explanation

Solution

cos⁡θ=x2+(l−x)2−a22x(l−x)=T\cos\theta = \frac{x^{2} + (l - x)^{2} - a^{2}}{2x(l - x)} = T

T=x2+l2+x2−2lx−a22lx−2x2T = \frac{x^{2} + l^{2} + x^{2} - 2lx - a^{2}}{2lx - 2x^{2}}

⇒ T=2x2−2lx+l2−a22lx−2x2T = \frac{2x^{2} - 2lx + l^{2} - a^{2}}{2lx - 2x^{2}} = −1+12(l2−a2xl−x2)- 1 + \frac{1}{2}\left( \frac{l^{2} - a^{2}}{xl - x^{2}} \right)

⇒ dtdx=−12(l−2x)(l2−a2)(lx−x2)2\frac{dt}{dx} = - \frac{1}{2}\frac{(l - 2x)\left( l^{2} - a^{2} \right)}{\left( lx - x^{2} \right)^{2}}=0 ; x=l2x = \frac{l}{2}

'θ' is maximum when x=l2x = \frac{l}{2} triangle is an isosceles.

So 'b' is correct