Question

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

Answer 1

Let $ABC$ be the given triangle.
Let $AB=c=4cm$ and
$AC=b=5cm$ and $\theta$ be the angle between $AB$ and $AC$ at time $t$.
Given, $\frac{d\theta }{dt}=0.06$ rad/sec.
Area of the triangle $=\frac{1}{2}bc\sin \theta =\frac{1}{2}\times 4\times 5\sin \theta$
$\Rightarrow A=10\sin \theta$
Differenting w.r.t. ${}^{\prime }{\theta }^{\prime }$ we get,
$\frac{dA}{dt}=10\cos \theta \frac{d\theta }{dt}$
When $\theta =\frac{\pi }{3},\frac{dA}{dt}=10\cos \frac{\pi }{3}\left(0.06\right)$
$=10\left(\frac{1}{2}\right)\left(0.06\right)=0.3m^2/sec$.
$\therefore$ the rate of increase in area $=0.3m^2/sec$.