Since this problem talks about rates of change, then the concept of calculus is very useful. But first, let's find at least two equations in order to solve this system. The first one is the area of a triangle written as
A = 1/2 ab sin θ, where a and b are the sides that from the angle θ. So, we substitute a=6 and b=10. That makes it: A = 1/2 (6)(10)sin θ = 30 sin θ
Now, you differentiate implicitly (both sides simultaneously) with respect to time.
dA/dt = 30 cosθ (dθ/dt)
We replace dθ/dt = 0.06 rad/s, as mentioned in the problem. Then, the rate of change of the area of the triangle when θ = π/3 rad with respect to time is
dA/dt = 30cos(π/3) (0.06) dA/dt = 1.8 m²/s
Therefore, the rate of change of the area of the triangle is 1.8 m² per second.
