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. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is π 3 rad.

Answer :

amna04352

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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