



 
Hi Gevork, The area of a parallelogram is the base times the height. In my diagram below the height $h$ is related to the side of length 8 and the angle $\theta$ in radians by $\sin \theta = \frac{h}{8}.$ Solving this expression for $h$ and substituting into the area expression for the area gives \[A(t) = 72 \sin \theta(t).\] I have written the area $A$ and the angle $\theta$ as functions of time $t$ in minutes as they are both changing with respect to time. You can now differentiate both sides with respect to $t$ to find $\large \frac{dA}{dt}$ as a function of $\theta$ and $\large \frac{d\theta}{dt.}$ The remaining challenge is that $\theta$ is in radians and $\large \frac{d\theta}{dt}$ is in radians per minute and you have an angle of 30 degrees which is changing at 2 degrees per minute. To evaluate $\large \frac{dA}{dt}$ at the correct time you will need to convert radians to degrees using the fact that $\pi \mbox{ radians } = 180 \mbox{ degrees }.$ I hope this helps,  


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