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
(x-8)^2 = (x^2+3x+9)+(x^2-5x+25)-2√((x^2+3x+9)(x^2-5x+25))cos(30)
which simplifies to the quartic equation
2x^4-40x^3+143x^2+120x-225=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 not-so-nice equation.

Posted by Jer on 2015-02-28 13:58:00