Question

Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 2 units, b = 4 units, and c = 6 units. Round your answer to two decimal places.)

Answer 1

Given a = 2 units, b = 4 units, and c = 6 units
Let x = measure of PB. This makes PA = 4 - x.
Let C = other point of the triangle with P and B. Let y = <PCB.
Let D = other point of the triangle with P and A. Let z = <PDA.
Notice that theta = y + z.
tan y = x/a = x/2
y =arctan(x/2)
tan z = (4 - x)/c = (4 - x)/6
z = arctan(4 - x)/6
Therefore, theta is a function of x, namely f(x) = arctan(x/2) + arctan(4 - x)/6
f(x)'s maximum occurs where f '(x) = 0.
f '(x) = 1/2(1/(1 + (x/2)^2) - 1/6(1/(1 + ((4 – x)/6)^2))
f'(x)=2/(x^2+4)-3/(x^2-8x+52)
f '(x) = 0 when
=>2/(x^2+4)-3/(x^2-8x+52)=0
x = 2(-4-√39)
x=-20.49
or x = 2(√39-4)
x=4.49
A negative length makes no sense so discard x = -20.49, leaving just
x = 4.49


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