Related Rates of a Triangle with Fixed Side Lengths

[Greywolfe1982]

Homework Statement

Two sides of a triangle of fixed length measure 12m and 15m. The angle between these two sides increases at a rate of 2 degrees per minute. When the angle between these two sides is 60 degrees, at what rate is the third side increasing?

Homework Equations

That's the problem.

The Attempt at a Solution

I'm not sure what equation I should differentiate - law of sines would give be division by zero (something over (da/dt)^2), and law of cosines would give me 0 = 0, as the length of the sides are not changing. What equation am I forgetting here, or am I going at this from completely the wrong way?

 

[fzero]

Take a more careful look at the law of cosines. Post your work if you are still confused.

 

[Dickfore]

Hint:

Assume the angle between the sides is a variable \alpha.

Express the third side in terms of the lengths of the two sides and the angle between them.

Differentiate with respect to t according to the chain rule, assuming \alpha = \alpha(t). This will give you the rate with which the length of the third side increases. You may substitute \alpha = 60^{\circ}.

Express \dot{\alpha} in rad/s instead of degrees/s.