A triangle has two constant sides of length 3 feet and 5 feet. the angle between these two sides is increasing at a rate of 0.1 radians per second. find the rate at which the area of the triangle is changing when the angle between the two sides is π/6.

Answer:The rate of change of the area is 0.65 feet²/secondStep-by-step explanation:* Lets explain how to solve the problem- A triangle has two constant sides of length 3 feet and 5 feet∵ The angle between them is Ф- The angle between these two sides is increasing at a rate of  0.1 radians per second∴ dФ/dt = 0.1 radians/second- The rule of the area of a triangle whose sides are a , b and the angle   between them is Ф is ⇒ A = 1/2 × a × b × sin Ф∵ The two sides of the triangle are 3 feet and 5 feet∵ The area of the triangle = 1/2 × 3 × 5 × sin Ф∴ A = 7.5 sin Ф- The rate of change of the area is dA/dt - Lets find the differentiation of the area with respect to angle Ф∵ A = 7.5 sin Ф- Remember the differentiation of sin Ф is cos Ф∴ dA/dФ = 7.5 cos Ф - The angle between the two sides is π/6∴ Ф = π/6∴ dA/dФ = 7.5 cos(π/6)- The rate of change of the area is dA/dt ∵ dA/dФ = 7.5 cos(π/6) ⇒ (1)∵ dФ/dt = 0.1 ⇒ (2)- Multiply (1) and (2) to find dA/dt∴ dA/dФ × dФ/dt = 7.5 cos(π/6) × 0.1 ⇒ (dФ up cancel dФ down)∴ dA/dt = 0.75 cos(π/6) = (3√3)/8 = 0.649519∴ dA/dt ≅ 0.65 feet²/second* The rate of change of the area is 0.65 feet²/second