Question
Construct a triangle using the given data: $PQ = 6.2\ cm, \angle P = 105^\circ $ and $\angle = 45^\circ $
✓
Answer
$PQ = 6.2\ cm, \angle P = 105^\circ $ and $\angle = 45^\circ $
Steps of Construction:
$1$. Draw a line segment $PQ = 6.2\ cm.$
$2$. With $P$ as centre, draw an arc meeting $PQ$ at $A.$
$3$. With $A$ as centre and same radius, cut the arc at $B$ and with $BQ$ as centre and same radius, cut the arc at $C.$
$4$. With $B$ and $C$ as centre, cut arcs and draw $PM$ perpendicular to $PQ.$
$5$. Bisect $\angle MPC$. Let $PX$ be the bisector. $PX$ makes an angle of $105^\circ $ with $PQ.$
$6$. With $Q$ as centre, draw an arc meeting $PQ$ at $S.$
$7$. With $S$ as centre and same radius, cut the arc at $T$ and with $T$ as centre and same radius, cut the arc at $U.$
$8.$ With $T$ and $U$ as centre, cut arcs and draw $QN$ perpendicular to $PQ.$
$9$. Bisect $\angle NQP.$ Let $QY$ be the bisector. $QY$ makes an angle of $45^\circ $ with $PQ.$
$10.$ Mark the point as $R$, where $PX$ and $QY$ cut each other.
Thus,$\text{PQR}$ is the required triangle.
Steps of Construction:
$1$. Draw a line segment $PQ = 6.2\ cm.$
$2$. With $P$ as centre, draw an arc meeting $PQ$ at $A.$
$3$. With $A$ as centre and same radius, cut the arc at $B$ and with $BQ$ as centre and same radius, cut the arc at $C.$
$4$. With $B$ and $C$ as centre, cut arcs and draw $PM$ perpendicular to $PQ.$
$5$. Bisect $\angle MPC$. Let $PX$ be the bisector. $PX$ makes an angle of $105^\circ $ with $PQ.$
$6$. With $Q$ as centre, draw an arc meeting $PQ$ at $S.$
$7$. With $S$ as centre and same radius, cut the arc at $T$ and with $T$ as centre and same radius, cut the arc at $U.$
$8.$ With $T$ and $U$ as centre, cut arcs and draw $QN$ perpendicular to $PQ.$
$9$. Bisect $\angle NQP.$ Let $QY$ be the bisector. $QY$ makes an angle of $45^\circ $ with $PQ.$
$10.$ Mark the point as $R$, where $PX$ and $QY$ cut each other.
Thus,$\text{PQR}$ is the required triangle.
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