Answer:
\[PQ=5\text{ }cm,\,\,\angle PQR=105{}^\circ ,\angle QRP=40{}^\circ \]
\[\because \]Sum of angles of a triangle is 180°
\[~\therefore \angle PQR+\angle QRP+\angle QPR=180{}^\circ \]
\[\therefore 105{}^\circ +40{}^\circ +\angle QPR=180{}^\circ \]
\[\therefore 145{}^\circ +\angle QPR=180{}^\circ \]
\[\angle QPR\text{ }=\text{ }180{}^\circ 145{}^\circ \]
\[\angle QPR\text{ }=\text{ }35{}^\circ \]
Steps of Construction:
(a) Draw a line segment PQ of 5 cm.
(b) From point P draw an angle of 35°, such that
\[\angle QPX=35{}^\circ \]
(c) From point Q draw an angle of 105°, such that
\[\angle PQY\text{ }=\text{ }150{}^\circ \]
(d) These lines PX and QY cut at R.
PQR is required triangle.
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