Use the attached figure to guide yourself through the explanation. In the figure:
[tex]a=21cm, b=36cm, \widehat{C}=\frac{\pi}{6}[/tex].
We know that:
[tex]\frac{da}{dt}=9,\frac{db}{dt}=-2[/tex]
Also remmeber that the area of a triangle is:
[tex]A=\frac{Base\times Height}{2}[/tex]
The hegiht is shown in the figure in red, the base is just the b side.
Using the angle C the area of the triangle is:
[tex]A=\frac{1}{2}ab\cos(\widehat{C})\implies2A=ab\cos(\widehat{C})[/tex]
Now taking the derivative of the area 2A with respect to t we have:
[tex]2\frac{dA}{dt}=b\cos(\widehat{C})\frac{da}{dt}+a\cos(\widehat{C})\frac{db}{dt}-ab\sin(\widehat{C})\frac{d\widehat{C}}{dt}[/tex]
Replacing with the right values:
[tex]\frac{dA}{dt}=0\\
\\
\cos({\widehat{C}})=\frac{\sqrt{3}}{2}\\ \\
\sin({\widehat{C}})=\frac{1}{2}\\ \\
[/tex]
yields:
[tex]0=36\frac{\sqrt{3}}{2}.9-21\frac{\sqrt{3}}{2}.2-\frac{1}{2}.21.36.\frac{d\widehat{C}}{dt}\implies \frac{d\widehat{C}}{dt}=\frac{141}{378}\sqrt{3}[/tex]
