Exercise with angle bisector: find x1, x2 explicitly as function of L, D and alpha

Kolski

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Nov 7, 2017
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Let's consider the triangle:

b.jpg

Data: L, D and alpha, L is angle bisector

| would like to find out lenght x1 and x2 explicitly as a function of L, D and alpha.

I tried:

https://en.wikipedia.org/wiki/Angle_bisector_theorem

x1/ (d-k) = x2/k

and then 2x law of cosines, but I get fifth degree polynolomial, which could not be solved analitically.

Can anyone help me ?

 
Also, we don't know what YOU know;
using triangle side lengths 3,5,7,

I must calculate x1 and x2, not angles.

are you able to calculate its angles?

I can calculate it only for very special cases.

For example, if alpha = pi/2, Pythagorean theorem is enough:

(D/2)^2 = L^2 + x^2

x1 = x2 = x

But I want find general formula for any alpha.
 
Sorry, don't know what you mean…
Introducing a coordinate system to help explain, I read it like this:

On a horizontal t-axis, point A is at (0,0) and point B is at (D,0). Line segment AB is the base of some triangle ABC, where point C is in Quadrant I.

Take a line segment (of length L) and position it so that it forms an acute angle alpha with the t-axis and is connected to the t-axis at a point E between points A and B. Let the coordinates of point E be (t,0).

Point C is then located at (t + L·cos[alpha], L·sin[alpha]).

For any alpha between zero and Pi/2, with appropriately-related L & D given, there is a location of E where angles ACE and BCE are equal.

I think Kolski would like to know the resulting lengths of sides a and b (labeled x1 and x2, in their diagram), expressed in terms of constants L & D and variable alpha.

It would be nice to know how L & D are related.
 
I played with this exercise, while waiting for an appointment to show up. I obtained expressions for Kolski's x1 and x2 (i.e., sides b and a, respectively, in my earlier post), but I couldn't simplify them much. Basing my approach on t may have gummed up the works!

x1 = sqrt[t^2 + 2·t·L·cos(α) + L^2]

x2 = sqrt[t^2 + 2·t·L·cos(α) + L^2 + D^2 - 2·D·t - 2·D·L·cos(α)]

where

t = -1/2·(-(2·L^2·sin(α)^2 - D^2·sin(α)^2 + D^2 + 2·sqrt(L^4·sin(α)^4 + L^2·sin(α)^4·D^2 - L^2·sin(α)^6·D^2))·D/(4·L^2·sin(α)^2 + D^2) - D·sin(α)^2 + D + 2·L·cos(α)·(2·L^2·sin(α)^2 - D^2·sin(α)^2 + D^2 + 2·sqrt(L^4·sin(α)^4 + L^2·sin(α)^4·D^2 - L^2·sin(α)^6·D^2))/(4·L^2·sin(α)^2 + D^2) - 2·L·cos(α))/((2·L^2·sin(α)^2 - D^2·sin(α)^2 + D^2 + 2·sqrt(L^4·sin(α)^4 + L^2·sin(α)^4·D^2 - L^2·sin(α)^6·D^2))/(4·L^2·sin(α)^2 + D^2) - 1 + sin(α)^2)

:shock:
 
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