In an equilateral triangle ABC on the side AC taken points P and Q such that AP=3, QC=5. (A>P>Q>C). It is known that angle PBQ=30.
Find the side of the triangle ABC
Call x the missing side.
By the law of cosines on triangle ABP
BP^2 = x^2  3x + 9
By the law of cosines on triangle CBQ
BQ^2 = x^2  5x + 25
Using these results and the law of cosines one more time on triangle PBQ
(x8)^2 = (x^2+3x+9)+(x^25x+25)2√((x^2+3x+9)(x^25x+25))cos(30)
which simplifies to the quartic equation
2x^440x^3+143x^2+120x225=0
which has solutions
x={2±√11.5, 1, 15}
Of which only the last is actually possible.
So
the side length is 15.
Incidentally, my first attempt was to find the missing angles using the law of sines a few times. I don't recommend it. It worked (to a point) but I needed to use a graph to solve a notsonice equation.

Posted by Jer
on 20150228 13:58:00 