Answer :
Answer:
1.8 m²/s
Step-by-step explanation:
A = ½(10)(12)sin(theta)
A = 60sin(theta)
dA/d(theta) = 60cos(theta)
At theta = pi/3,
dA/d(theta) = 60cos(pi/3) = 30
dA/dt = dA/d(theta) × d(theta)/dt
dA/dt = 30 × 0.06
dA/dt = 1.8 m²/d
The area of the triangle is increasing at the rate of [tex]1.8m^2/d[/tex]
We have two sides of a triangle are 10 m and 12 m in length and the angle between them is increasing at a rate of 0.06 rad/s.
Area of the triangle = [tex]\frac{1}{2} absin\theta[/tex]
Differentiating it w.r.t [tex]\theta[/tex]
[tex]\frac{dA}{d\theta} = \frac{1}{2}absin(theta)\\\frac{dA}{d\theta} = \frac{1}{2}(10)(12)sin(theta)\\\frac{dA}{d\theta} = 60cos(\theta)[/tex]
At [tex]\theta =\frac{\pi}{3}[/tex],
[tex]\frac{dA}{d\theta} = 60cos(\frac{\pi}{3} ) \\= 30\\\frac{dA}{dt} =\frac{dA}{d\theta}. \frac{d\theta}{dt}\\\frac{dA}{dt} = 30(0.06)\\\frac{dA}{dt} = 1.8 m^2/d[/tex]
Therefore the area of the triangle is increasing at the rate of [tex]1.8m^2/d[/tex]
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