Question
Two vertical poles respectively 1 meter and 9 meters high are 6 meters apart. How far from the foot of the shorter pole where the line segment joining the tops of the poles subtends the greatest
angle? (please show the compution)
angle? (please show the compution)
Answers
Mark a point P which is x meters from the shorter pole.
Draw the lines from P to the tops of the poles.
Now you have three angles, A,B,C, and you want to know where P must be so that angle B is a maximum.
A+B+C = π
so
B = π - (A+C)
tanB = -tan(A+C) = (tanA+tanC)/(tanA tanC - 1)
tanB = (1/x + 9/(6-x))/((1/x)(9/(6-x))-1) = (8x+6)/(x-3)^2
sec^2B dB/dx = -4(2x+9)/(x-3)^3
dB/dx = -4(2x+9)/(x-3)^3 * 1/(1+((8x+6)/(x-3)^2)^2)
dB/dx = -4(2x+9)(x-3)^2 / (x^2+1)(x^2-12x+117)
As expected, P is halfway between the feet of the two poles, at x=3.
Draw the lines from P to the tops of the poles.
Now you have three angles, A,B,C, and you want to know where P must be so that angle B is a maximum.
A+B+C = π
so
B = π - (A+C)
tanB = -tan(A+C) = (tanA+tanC)/(tanA tanC - 1)
tanB = (1/x + 9/(6-x))/((1/x)(9/(6-x))-1) = (8x+6)/(x-3)^2
sec^2B dB/dx = -4(2x+9)/(x-3)^3
dB/dx = -4(2x+9)/(x-3)^3 * 1/(1+((8x+6)/(x-3)^2)^2)
dB/dx = -4(2x+9)(x-3)^2 / (x^2+1)(x^2-12x+117)
As expected, P is halfway between the feet of the two poles, at x=3.
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