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.)

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,
[tex]\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)[/tex]
and
[tex]\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)[/tex]

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

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

To maximixe angle θ, the differentiation of θ with respect to x must be equal to zero.
i.e.
[tex] \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} [/tex]

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

Thus
[tex]x=9-6 \sqrt{2}=0.5147[/tex]

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

ahirohit963 ahirohit963 · Answer 2 · 2021-10-27T11:14:38+02:00

The distance from A of point P, so that angle θ is maximum is 0.51.

Given :

a = 4 units

b = 5 units

c = 9 units

Line segment AB.

Solution :

Let AB = x and let the angle opposite side 'a' be [tex]\alpha[/tex] and the angle opposite side 'c' be [tex]\beta[/tex].

From the diagram angle [tex]\alpha[/tex] is given by:

[tex]\rm tan\alpha = \dfrac{a}{b-x}[/tex]

[tex]\rm \alpha = tan^-^1 (\dfrac{a}{b-x})[/tex]

From the diagram angle [tex]\beta[/tex] is given by:

[tex]\rm tan\beta = \dfrac{c}{x}[/tex]

[tex]\rm \beta = tan^-^1 (\dfrac{c}{x})[/tex]

From the diagram angle [tex]\theta[/tex] is given by:

[tex]\rm \theta = 180^\circ-tan^-^1(\dfrac{4}{5-x})-tan^-^1(\dfrac{9}{x})[/tex] --- (1)

To maximize the angle [tex]\theta[/tex], differentiate equation (1) with respect to x we get,

[tex]\rm \dfrac{d\theta }{dt} = 0 -\dfrac{4}{x^2-10x+41}+\dfrac{9}{x^2+81}=0[/tex]

[tex]4(x^2+81) = 9(x^2-10x +41)[/tex]

[tex]5x^2-90x+45=0[/tex]

[tex]x^2-18x+9=0[/tex]

[tex]x = \dfrac{18\pm\sqrt{18^2-(4\times 9)} }{2}[/tex]

[tex]x=9\pm6\sqrt{2}[/tex]

[tex]x = 9-6\sqrt{2} = 0.51[/tex]

The distance from A of point P, so that angle [tex]\theta[/tex] is maximum is 0.51.

For more information, refer the link given below

https://brainly.com/question/22108412