Two sides of a triangle are 15 and 20 feet in length. How fast is the third side increasing if the angle between the given sides is 60 degrees and is increasing at a rate of 2 degrees per second.
To find the rate of change of the third side, we need to use the law of cosines to relate the sides and angles of a triangle. The law of cosines states that for a triangle with sides lengths a, b, and c, and angle C opposite side c:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, we have two sides with lengths 15ft and 20ft, and an angle of 60 degrees. Let's denote the third side's length as c and the angle between the given sides as C.
Given:
a = 15 ft
b = 20 ft
C = 60 degrees
We want to find dc/dt (the rate at which the third side is changing) when dC/dt (the rate at which the angle is changing) is 2 degrees per second.
To do this, we'll take the derivative of the law of cosines with respect to time:
2c * dc/dt = -2ab * sin(C) * dC/dt
Plugging in the given values:
2c * dc/dt = -2(15 ft)(20 ft) * sin(60 degrees) * (2 degrees/second)
Since sin(60 degrees) is equal to sqrt(3)/2, we have:
2c * dc/dt = -2(15 ft)(20 ft) * (sqrt(3)/2) * (2 degrees/second)
Simplifying:
2c * dc/dt = -30 ft * sqrt(3) * (2 degrees/second)
Dividing both sides by 2c:
dc/dt = (-30 ft * sqrt(3) * (2 degrees/second)) / (2c)
Substituting the known values for c:
dc/dt = (-30 ft * sqrt(3) * (2 degrees/second)) / (2 * c)
Now we can calculate dc/dt by substituting c = √(a^2 + b^2 - 2ab * cos(C)):
dc/dt = (-30 ft * sqrt(3) * (2 degrees/second)) / (2 * √((15 ft)^2 + (20 ft)^2 - 2(15 ft)(20 ft) * cos(60 degrees)))
Simplifying further:
dc/dt = (-30 ft * sqrt(3) * (2 degrees/second)) / (2 * √(225 ft^2 + 400 ft^2 - 600 ft^2 * 1/2))
dc/dt = (-30 ft * sqrt(3) * (2 degrees/second)) / (2 * √(225 ft^2 + 400 ft^2 - 300 ft^2))
dc/dt = (-30 ft * sqrt(3) * (2 degrees/second)) / (2 * √(325 ft^2))
dc/dt = (-30 ft * sqrt(3) * (2 degrees/second)) / (2 * 5 ft)
Simplifying finally:
dc/dt = -3 ft * sqrt(3) * (2 degrees/second)
Therefore, the rate at which the third side is increasing is approximately -6sqrt(3) ft/sec.