Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at a rate of 0.06 rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed lengths is π/3.

Solution:

Let ABC be the given triangle

Where AB = 4 m and AC = 5 m

And θ be the angle the side AB and AC .

Given, the increasing rate of the angle with respect of time dθ/dt = 0.06 rad/sec.

Area of the triangle is given by 1/2 base × height

=> 1/2 (AB)(AC)

=> 1/2 × 4 × 5sinθ

=> A = area = 10 sinθ

Differentiating w.r.t θ we get

=> dA/dt = 10 sinθ dθ/dt

=> dA/dt = 10 cosθ dθ/dt

Now when θ = π/3

dA/dt = 10 cos π/3 (0.06)

=> 10(1/2)(0.06)

=> 0.3 m²/sec.

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Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at a rate of 0.06 rad/sec. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed lengths is π/3.

Summary:

The rate at which the area of the triangle is increasing, when the angle between the sides of fixed lengths is π/3 and the angle between them is increasing at a rate of 0.06 rad/sec is 0.3 m²/sec.