Two sides of a triangle are 8 m and 10 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.

Respuesta :

Answer:

The area of the triangle is increasing at a rate 1.2 m²/s.

Step-by-step explanation:

Given sides of a triangle are 8 m and 10 m in length.

The angle between the sides of the triangle is increasing at a rate 0.06 rad/s.

It means [tex]\frac{d\theta}{dt}=0.06[/tex]

If the sides of a triangle are a and b and θ is the angle of the sides.

The area of the triangle is A = [tex]\frac{1}{2}ab sin\theta[/tex]

Here, a = 8 m and b = 10 m

[tex]A= \frac{1}{2}\times 8 \times 10 sin \theta[/tex]

[tex]\Rightarrow A = 40 sin \theta[/tex]

Differentiating with respect to t

[tex]\frac{dA}{dt}= 40 cos\theta \frac{d\theta}{dt}[/tex]

putting the value of [tex]\frac{d\theta}{dt}[/tex]

[tex]\Rightarrow \frac{dA}{dt}|_{\theta=\frac{\theta}{3}}= 40 cos\frac{\pi}{3} \times 0.06[/tex]

                [tex]= 40 \times \frac{1}{2} \times 0.06[/tex]

                =1.2 m²/s

The area of the triangle is increasing at a rate 1.2 m²/s.