In an isosceles trapezium,the length of one o | vedclass

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(C) Let the parallel sides be $a = 30$ and $b = 30 + 2(30 \cos 	heta) = 30 + 60 \cos 	heta$. The height of the trapezium is $h = 30 \sin 	heta$.The

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$\frac{\pi}{3}$

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(C) Let the parallel sides be $a = 30$ and $b = 30 + 2(30 \cos 	heta) = 30 + 60 \cos 	heta$. The height of the trapezium is $h = 30 \sin 	heta$.The area $A$ of the trapezium is given by:$A = \frac{1}{2}(a + b)h = \frac{1}{2}(30 + 30 + 60 \cos 	heta)(30 \sin 	heta)$$A = \frac{1}{2}(60 + 60 \cos 	heta)(30 \sin 	heta) = 900(1 + \cos 	heta) \sin 	heta = 900(\sin 	heta + \sin 	heta \cos 	heta) = 900(\sin 	heta + \frac{1}{2} \sin 2	heta)$.To maximize the area,differentiate with respect to $	heta$ and set to $0$:$\frac{dA}{d	heta} = 900(\cos 	heta + \cos 2	heta) = 0$.Since $\cos 2	heta = 2 \cos^2 	heta - 1$,we have:$2 \cos^2 	heta + \cos 	heta - 1 = 0$.$(2 \cos 	heta - 1)(\cos 	heta + 1) = 0$.This gives $\cos 	heta = \frac{1}{2}$ or $\cos 	heta = -1$.Since $	heta$ is an angle in a triangle,$	heta = \frac{\pi}{3}$ (as $	heta = \pi$ is not possible).Thus,the smallest angle is $\frac{\pi}{3}$.

In an isosceles trapezium,the length of one of the parallel sides and the lengths of the non-parallel sides are all equal to $30$. In order to maximize the area of the trapezium,the smallest angle should be:

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