Step 1:

Let $OA,OB$ be the sides of $\Delta\; OA$ and $OB$ where $OA=4m,OB=5m \;and \;\angle AOB=\theta$

The area of the triangle at time 't' is

$A=\large\frac{1}{2}$$OA \times OB \sin \theta$

$\quad=\large\frac{1}{2}$$ \times 4 \times 5 \sin \theta$

$\quad=10 \sin \theta$

Step 2:

The angle $\theta$ is increasing at the rate of $0.06^c/sec$

(ie) $\large\frac{d \theta}{dt}$$=0.06^c/sec$

Step 3:

The rate of change of the area is

$\large\frac{dA}{dt}=$$10 \cos \theta \large\frac{d \theta}{dt}$

Step 4:

When $\theta=\large\frac{\pi}{3} $ and $\large\frac{d\theta}{dt}$$=0.06$

$\large\frac{dA}{dt}$$=10 \cos \large\frac{\pi}{3} $$\times 0.06$

$\quad=10 \times \large\frac{1}{2}$$ \times 0.06$

$\quad=0.3 m^2 /sec$

The area is increasing at $0.3 m^2 /sec$