Two sides of a triangle have lengths 11 m and 14 m. The angle between them is increasing at a rate of 2°/min.

Question

Two sides of a triangle have lengths 11 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.)

P.S. The answers are NOT (-3.8, 20.892)

Answer 1

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Perhaps the angle needs to be in radians, not degrees...

maybe you must plug in dC/dt = pi/90 instead of 2

Answer 2

Let fixed sides of triangle be a = 11, b = 14,

Let unknown and increasing side be c

Let the angle between a and b be x With x' = 2 deg/min.

NOTE: In these equations, use radians, NOT degrees.

NOTE: Change degrees to radians by multiplying by [pi/180]

Use Law of Cosines to find c when x = 60 degrees

c^2 = a^2 + b^2 - 2ab*cos(60)

c^2 = 11^2 + 14^2 - 2(11)(14)*(1/2)

c^2 = 163

c = sqrt(163)

Use Law of Cosines to show relationship between side c

and the increasing angle.

c^2 = a^2 + b^2 - 2ab*cos(x)

d/dt [c^2] = d/dt [a^2 + b^2 - 2ab*cos(x) ]

2cc' = 0 + 0 - 2(a)(b)*( -sin(x) x' )

2*sqrt(163)c' = 0 + 0 - 2(11)(14)*( -sin(60) x' )

2*sqrt(163)c' = 2(11)(14)*( sqrt(3)/2 )*(2)*[pi/180]

2*sqrt(163)c' = (308)*( sqrt(3)/2 )*(2)*[pi/180]

2*sqrt(163)c' = (77/45)*(pi)*sqrt(3)

c' = [ (77/45)*(pi)*sqrt(3) ] / [ 2*sqrt(163) ]

c' = [ (77/90)*(pi)*sqrt(3) ] / [ sqrt(163) ]

c' = 77*(pi)*sqrt(489) / 14670

c' = .3646405 m/min.