Two sides of a triangle have lengths 13 m and 17 m. The angle between them is increasing at a rate of 2°/min.

Question

How fast (in m/min) is the length of the third side increasing when the angle between the sides of fixed length is 60°?

Answer 1

b = 13

c= 17

A = 60°

sin A = √3/2

cos A = 1/2

dA/dt = 4°/min = 4 * π/180 = π/45 radians/min

The cos law is a = √(b^2 + c^2 - 2bc cos A)

Thus da/dA = bc sin A/(√b^2 + c^2 - 2bc cos A)

and da/dt = da/dA dA/dt =

bc sin A/√(vb2 + c^2 - 2bc cos A) * π/45/min =

13 * 17 * √3/2/√(13^2 + 17^2 - 2*13*17/2) * π/45/min =

221 * √3/2/√237 * π/45/min or

221√3//(90 √237) π/min or

.8679/min