Question

Two sides of a triangle are $8$ m and $5$ m in length. The angle between them is increasing at the rate $0.08$ rad/sec. When the angle between the sides is $\frac{\pi}{3}$, the rate at which the area of the triangle is increasing is:

• A. $0.4 \text{ m}^2/\text{sec}$
• B. $0.8 \text{ m}^2/\text{sec}$
• C. $0.6 \text{ m}^2/\text{sec}$
• D. $0.04 \text{ m}^2/\text{sec}$
• E. $0.08 \text{ m}^2/\text{sec}$

Hint:

In related rates problems involving triangles, if the side lengths are fixed, the only variable affecting the area is the angle. This makes the derivative much simpler than using Heron's formula or base/height methods.

Answer 1

The Correct Option is B

Concept: The area $A$ of a triangle with sides $a$ and $b$ and included angle $\theta$ is given by $A = \frac{1}{2}ab \sin \theta$. To find the rate of change of area, we differentiate with respect to time $t$.

Step 1: Set up the area derivative.
Given $a = 8$, $b = 5$, and $A = \frac{1}{2}ab \sin \theta$: \[ A = \frac{1}{2}(8)(5) \sin \theta = 20 \sin \theta \] Differentiating both sides with respect to time $t$: \[ \frac{dA}{dt} = 20 \cos \theta \frac{d\theta}{dt} \]

Step 2: Identify the rates and values at the instant.

• Rate of change of angle: $\frac{d\theta}{dt} = 0.08$ rad/sec.
• Angle at the instant: $\theta = \frac{\pi}{3}$.

Step 3: Calculate the rate of change of area.
Substitute the values into the derivative equation: \[ \frac{dA}{dt} = 20 \cos\left(\frac{\pi}{3}\right) \times 0.08 \] Since $\cos(\pi/3) = 1/2$: \[ \frac{dA}{dt} = 20 \times \frac{1}{2} \times 0.08 = 10 \times 0.08 = 0.8 \text{ m}^2/\text{sec} \]