Draw a triangle \(PQR\) of base \(PQ = 6.8 \ cm\), vertical angle is \(60^{\circ}\) and the bisector of the vertical angle meets the base at \(D\) where \(PD = 5.2 \ cm\).
\(Step-1:\) Draw a line segment \(PQ\) of length \(\ cm\).
\(Step-2:\) At \(P\), draw \(PE\) such that \(\angle QPE =\) \(^{\circ}\).
\(Step-3:\) At \(P\), draw \(PF\) such that \(\angle EPF =\) \(^{\circ}\).
\(Step-4:\) Draw the to \(PQ\), which intersects \(PF\) at \(O\) and \(PQ\) at \(G\).
\(Step-5:\) With \(O\) as centre and \(OP\) as radius draw a .
\(Step-6:\) From \(P\) mark an of \(5.2 \ cm\) on \(PQ\) at \(D\).
\(Step-7:\) The intersects the circle at \(I\). Join \(ID\).
\(Step-8:\) \(ID\) produced meets the at \(R\). Now, join \(PR\) and \(RQ\). Then, \(\triangle PQR\) is the required triangle.