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The base of an isosceles triangle is 8 feet long. If the altitude is 6 feet long and is increasing 3 inches per minute, at what rate are the base angles changing?

Hi Christine,

The first thing we have to do is relate the base angle to the altitude. Consider half of the isosceles triangle to one side and you have a right-angled triangle with a 4 feet long base. If the base angle is θ and the altitude H, then the relationship between them is given by;

tanθ = H/4

To find the rates of change we differentiate both sides with respect to time t.

d(tanθ)/dt = d(H/4)/dt

d(tanθ)/dθ . dθ/dt = (1/4)dH/dt

secēθ dθ/dt = (1/4)dH/dt

And we immediately see that the rate of change of the base angle dθ/dt depends on the value of dH/dt and secēθ at that time. So, at H = 6 feet

tanθ = 6/4 = 1.5

We can use this to find secēθ from the trigonometric identity

secēθ = 1 + tanēθ

= 1 + 2.25

= 3.25

Therefore, at the moment when the altitude is 6 feet and increasing at 3 inches/minute

dH/dt = 3 inches/minute = 0.25 feet/minute

3.25 dθ/dt = (1/4)(0.25)

dθ/dt = (0.25/4)/3.25

= 0.019 radians/minute

Be careful not to write the unit for the angle in degrees since d(tanθ)/dθ is only secēθ when the angle unit is in radians.

Regards

Calculus

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