Question

Two sides of a triangle have lengths 12 m and 14 m. The angle between them is increasing at a rate of 2°/min. How fast is the length of the third side increasing when the angle between the sides of fixed length is 60°? (Round your answer to three decimal places.) webassig n

Answer 1

Answer:

1.692 m/min

Step-by-step explanation:

Let $\theta$ be the angle between the two sides and x be the length of the third side. By cosine rule,

$x^2 = 12^2+14^2-2\times12\times14\cos\theta = 340 - 336\cos\theta$

$x= \sqrt{340 - 336\cos\theta}$

We differentiate x with respect to $\theta$ by applying chain rule.

$\dfrac{dx}{d\theta} = \dfrac{336\sin\theta}{2\sqrt{340 - 336\cos\theta}} = \dfrac{168\sin\theta}{\sqrt{340 - 336\cos\theta}}$

Rate of change of $\theta$ is 2

$\dfrac{\theta}{dt} = 2$

Rate of change of x is

$\dfrac{dx}{dt} = \dfrac{dx}{d\theta}\times\dfrac{d\theta}{dt}$

$\dfrac{dx}{dt} = \dfrac{168\sin\theta}{\sqrt{340 - 336\cos\theta}} \times2=\dfrac{336\sin\theta}{\sqrt{340 - 336\cos\theta}}$

At 60°,

$\dfrac{dx}{dt} = \dfrac{336\sin60}{\sqrt{340 - 336\cos60}} = 1.692 \text{ m/min}$