P, Q and R are points such that ∠PQR = 75° and PR = 9 cm. The line PQ has been drawn for you.

1. Draw the ray from $Q$ making $75^\circ$ with $QP$.

At $Q$, use a protractor to measure $75^\circ$ from the line $QP$. Draw a ray from $Q$ in the correct direction for $QR$ so that
$\angle PQR=75^\circ.$

2. Locate $R$ using the condition $PR=9\text{ cm}$ and complete the triangle.

Set the compass to $9\text{ cm}$. With centre $P$, draw an arc to meet the ray from $Q$ at $R$. Join $P$ to $R$ and $Q$ to $R$.

Then $PR=9\text{ cm}$ and $R$ lies on the ray making $75^\circ$ with $QP$.

1. Draw equal-radius arcs from $P$ and $Q$ to locate two points equidistant from $P$ and $Q$.

Set the compass to a radius greater than half of $PQ$.

With centre $P$, draw arcs above and below $PQ$.

Using the same compass radius, with centre $Q$, draw arcs cutting the first pair of arcs above and below $PQ$.

2. Join the two arc intersection points to form the perpendicular bisector.

Use a ruler to draw the straight line through the two arc intersection points.

This line is the perpendicular bisector of $PQ$.

1. Construct a line from $Q$ that bisects $\angle PQR$.

With centre $Q$, draw an arc cutting $QP$ and $QR$ at two points. Using the same compass radius from these two points, draw arcs that intersect inside $\angle PQR$. Join $Q$ to this intersection point with a ruler. This line is the angle bisector of $\angle PQR$.