Two forces of magnitude p and 2q are inclined at an angle 150°. If the resultant force is perpendicular to p, show that p = √3q

Sol: The 2 forces magnitude are p,2q and let resultant be r.Now r^2=p^2+4q^2-2pq(3)^1/2,Now the three form right angle triangle where hypotenuse is 2q.So 4q^2=r^2+p^2 or 4q^2=2p^2+4q^2-2pq(3)^1/2 or 0=2p^2-2pq(3)^1/2 or 2pq(3)^1/2=2p^2 finally; p=q(3)^/12