Question

One side of a triangle is increasing at a rate of 9 cm/s and a second side is decreasing at a rate of 2 cm/s. if the area of the triangle remains constant, at what rate does the angle between the sides change when the first side is 21 cm long, the second side is 36 cm, and the angle is π/6? (round your answer to three decimal places.)

Answer 1

Use the attached figure to guide yourself through the explanation. In the figure:
$a=21cm, b=36cm, \widehat{C}=\frac{\pi}{6}$.
We know that:
 $\frac{da}{dt}=9,\frac{db}{dt}=-2$
Also remmeber that the area of a triangle is:
$A=\frac{Base\times Height}{2}$
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:
$A=\frac{1}{2}ab\cos(\widehat{C})\implies2A=ab\cos(\widehat{C})$
Now taking the derivative of the area 2A with respect to t we have:
$2\frac{dA}{dt}=b\cos(\widehat{C})\frac{da}{dt}+a\cos(\widehat{C})\frac{db}{dt}-ab\sin(\widehat{C})\frac{d\widehat{C}}{dt}$

Replacing with the right values:
$\frac{dA}{dt}=0\\ \\ \cos({\widehat{C}})=\frac{\sqrt{3}}{2}\\ \\ \sin({\widehat{C}})=\frac{1}{2}$
yields:
$0=36\frac{\sqrt{3}}{2}.9-21\frac{\sqrt{3}}{2}.2-\frac{1}{2}.21.36.\frac{d\widehat{C}}{dt}\implies \frac{d\widehat{C}}{dt}=\frac{141}{378}\sqrt{3}$