Rate of Change in Area of Triangle with Fixed Sides

[karush]

$29.$ Two sides of a triangle are $4 \, m$ and $5 \, m$ in length and the angle between them is increasing at a rate of $\frac{0.06 \, rad}{s}$
Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length. $\frac{\pi}{3}$

$\displaystyle
A=\frac{1}{2}(4)(5\sin{\theta})\\
\text{differentiate}\\
\displaystyle
\frac{dA}{ds}
=10\cos{\theta}\frac{d\theta}{ds}\\
\displaystyle\theta=\frac{\pi}{3}
\text{ and }
\displaystyle
\frac{d\theta}{ds}
=\frac{0.6 rad}{s}\\
\text{so far??} $

 

[Prove It]

$29.$ Two sides of a triangle are $4 \, m$ and $5 \, m$ in length and the angle between them is increasing at a rate of $\frac{0.06 \, rad}{s}$
Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length

$\displaystyle
A=\frac{1}{2}(4)(5\sin{\theta})\\
\text{differentiate}\\
\displaystyle
\frac{dA}{ds}
=10\cos{\theta}\frac{d\theta}{ds}\\
\displaystyle\theta=\frac{\pi}{3}
\text{ and }
\displaystyle
\frac{d\theta}{ds}
=\frac{0.6 rad}{s}\\
\text{so far??} $

Are you told that the angle is pi/3?

 

[karush]

Yes $\theta$ is $\frac{\pi}{3}$
the book answer is $\frac{0.3m^2}{s}$

So $\d{A}{s}
=10\left(\frac{1}{2}\right)\frac{0.6\pi}{s}
=9.4248 \\$
Then $\d{A}{s}\approx\frac{0.3m^2}{s}$
kinda ??