Construct a triangle PQR with Q=90 and R=60 QR = 6 cm.Also construct the angle bisector of the angle p. (Steps of construction and justification)

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STEPS OF CONSTRUCTION :
$$\begin{gathered} \left(1\right). \mathrm{Draw} \mathrm{a} \mathrm{line} \mathrm{segment} \mathrm{QR} = 6 \mathrm{cm} \\ \left(2\right). \mathrm{At} \mathrm{Q}, \mathrm{draw} \angle \mathrm{XQR} = 90°. \\ \left(3\right). \mathrm{At} \mathrm{R}, \mathrm{draw} \angle \mathrm{YRQ} = 60°, \mathrm{that} \mathrm{intersects} \mathrm{ray} \mathrm{QX} \mathrm{at} \mathrm{P}. \\ \left(4.\right) \mathrm{We} \mathrm{get} \mathrm{PQR} \mathrm{as} \mathrm{the} ∆ \\ \left(5\right). \mathrm{With} \mathrm{P} \mathrm{as} \mathrm{centre}, \mathrm{draw} \mathrm{an} \mathrm{arc} \mathrm{of} \mathrm{convenient} \mathrm{radius} \mathrm{that} \mathrm{intersects} \mathrm{PR} \mathrm{at} \mathrm{L} \\ \mathrm{and} \mathrm{PQ} \mathrm{at} \mathrm{M}. \\ \left(6\right). \mathrm{With} \mathrm{M} \mathrm{as} \mathrm{centre} \mathrm{draw} \mathrm{an} \mathrm{arc} \mathrm{of} \mathrm{same} \mathrm{radius}. \\ \left(7\right). \mathrm{With} \mathrm{L} \mathrm{as} \mathrm{centre} \mathrm{draw} \mathrm{an} \mathrm{arc} \mathrm{of} \mathrm{same} \mathrm{radius} \mathrm{that} \mathrm{intersects} \mathrm{the} \mathrm{prevoius} \mathrm{arc} \mathrm{at} \mathrm{O}. \\ \left(8\right). \mathrm{Join} \mathrm{PO} \mathrm{and} \mathrm{produce} \mathrm{it} \mathrm{to} \mathrm{point} \mathrm{N} \mathrm{on} \mathrm{QR}. \\ \left(9\right). \mathrm{Hence} \mathrm{PN} \mathrm{is} \mathrm{the} \mathrm{bisector} \mathrm{of} \angle \mathrm{QPR}. \end{gathered}$$
 

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