Let x be the length of one of the equal sides of an isosceles triangle, and let $θ$ be the angle between them, If x is increasing at the rate $\frac{1}{12} m / h$ and $θ$ is increasing at the rate of $\frac{π}{180}$ radians/h then the rate in $m^{2} / h$ at which the area of the triangle is increasing when $x = 12 m and θ = π / 4$

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$\sqrt{3} (\frac{π}{5} + \frac{1}{2})$

$(\frac{π}{5} + \frac{1}{2})$

$\sqrt{2} (\frac{π}{5} + \frac{1}{2})$

$\sqrt{2} (\frac{π}{5} + \frac{1}{5})$

answer is B.

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Detailed Solution

$A = \frac{1}{2} x^{2} \sin θ or 2 A = x^{2} \sin θ$

$or 2 \frac{d A}{d t} = x^{2} \cos θ \frac{d θ}{d t} + \sin θ 2 x \frac{d x}{d t}$

$or 2 \frac{d A}{d t} = (144) (\frac{1}{\sqrt{2}}) \frac{π}{180} + \frac{1}{\sqrt{2}} \times 2 \times 12 \times \frac{1}{12}$

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