Question

Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 4 units, b = 5 units, and c = 9 units. Round your answer to two decimal places.)

Answer 1

From the figure, let the distance of point P from point A on line segment AB be x and let the angle opposite side a be M and the angle opposite side c be N.

Using pythagoras theorem,
$\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)$
and
$\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)$

Angle θ is given by
$\theta=180-M-N \\ \\ =180-\tan^{-1}\left(\frac{a}{b-x}\right)-\tan^{-1}\left(\frac{c}{x}\right)$

Given that a = 4 units, b = 5 units, and c = 9 units, thus
$\theta=180-\tan^{-1}\left(\frac{4}{5-x}\right)-\tan^{-1}\left(\frac{9}{x}\right)$

To maximixe angle θ, the differentiation of θ with respect to x must be equal to zero.
i.e.
$\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\ \\ -4(x^2+81)+9(x^2-10x+41)=0 \\ \\ -4x^2-324+9x^2-90x+369=0 \\ \\ 5x^2-90x+45=0 \\ \\ x^2-18x+9=0 \\ \\ x=9\pm6 \sqrt{2}$

Given that x is a point on line segment AB, this means that x is a positive number less than 5.

Thus
$x=9-6 \sqrt{2}=0.5147$

Therefore, The distance from A of point P, so that angle θ is maximum is 0.51 to two decimal places.