Construction a triangle from this conditions

Posted February 17th, 2008 by alpha

How to construct a triangle PQR in which QR=6cm,angle Q=60 degree and PQ-PR=2cm.

Comments

On February 19th, 2008 Structure says:

From cosine theorem

$a^2=b^2+c^2-2bc \cos A$

we have for PR=x, PQ=x+2

$x^2=6^2+(x+2)^2-2*6(x+2)\cos\frac{\pi}{3}$

$x^2=36+x^2+4x+4-12(x+2)\frac{1}{2}$

$x^2=36+x^2+4x+4-6(x+2)$

$2x=28$

$PR=x=14$

$PQ=x+2=16$