Question

A triangle has corners at points A, B, and C. Side AB has a length of `15`. The distance between the intersection of point A's angle bisector with side BC and point B is `12`. If side AC has a length of `36`, what is the length of side BC?

Answer 1

40.8

Explanation:

I may have misinterpreted the question, but using the wonderful power of MS Paint:

Let D be the point at which the bisector of A meets BC.
Let `BC=x`
`:. DC=x-12`
Let `BhatAC=2theta`
`:.BhatAD=DhatAC`

Let `BhatDA=alpha`
(I know I drew it wrong on the diagram, but) `:.AhatDC=180-alpha` (even though I wrote `alpha`. Just ignore that bit...)

from the sine rule in `triangleABD`:

`sintheta/12=sinalpha/15`
`sinalpha=(15sintheta)/12`

from `triangleACD`:

`sintheta/(x-12)=sin(180-alpha)/36`
`sin(180-alpha)=(36sintheta)/(x-12)`

From the graph `y=sinx` or from looking at the unit circle, `sinalpha=sin(180-alpha)`

`sinalpha=(36sintheta)/(x-12)`

But `sinalpha=(15sintheta)/12`

`(36cancelsintheta)/(x-12)=(15cancelsintheta)/12`

`36/(x-12)=(5cancel15)/(4cancel12)`

`5(x-12)=36*4`
`5x-60=144`
`5x=204`
`x=40.8`

Answer 2

Length of side BC = 40.8

Explanation:

Let the point where the angle bisector intersects with
side BC be D

`"using the "color(blue)"angle bisector theorem"`

`(AB)/(AC)=(BD)/(DC)`

`15 / 36 = 12 / (DC)`

`DC = (12*36) / 15 = 28.8`

`BC = BD+DC= 12+28.8 =40.8`