In the diagram below, QRT is a straight line. If angle PTR = 90°, angle PRT = 60° and angle PQR = 30° and (PQ) = 6 extsqrt(3) cm. Calculate (RT).

Since QRT is a straight line, $\angle PRQ$ and $\angle PRT$ are supplementary angles.

$$\angle PRQ = 180^\circ - \angle PRT$$ $$\angle PRQ = 180^\circ - 60^\circ$$ $$\angle PRQ = 120^\circ$$

The sum of angles in $\triangle PQR$ is $180^\circ$.

$$\angle QPR = 180^\circ - \angle PQR - \angle PRQ$$ $$\angle QPR = 180^\circ - 30^\circ - 120^\circ$$ $$\angle QPR = 30^\circ$$

Since $\angle PQR = \angle QPR = 30^\circ$, $\triangle PQR$ is an isosceles triangle, meaning $PR = QR$.
Using the Sine Rule:

$$\frac{PR}{\sin(\angle PQR)} = \frac{PQ}{\sin(\angle PRQ)}$$ $$\frac{PR}{\sin(30^\circ)} = \frac{6\sqrt{3}}{\sin(120^\circ)}$$

We know $\sin(30^\circ) = \frac{1}{2}$ and $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$.

$$PR = \frac{6\sqrt{3} \times \sin(30^\circ)}{\sin(120^\circ)}$$ $$PR = \frac{6\sqrt{3} \times \frac{1}{2}}{\frac{\sqrt{3}}{2}}$$ $$PR = \frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}}$$ $$PR = 3\sqrt{3} \times \frac{2}{\sqrt{3}}$$ $$PR = 6 \text{ cm}$$

In $\triangle PRT$, we have $\angle PTR = 90^\circ$, $\angle PRT = 60^\circ$, and the hypotenuse $PR = 6 \text{ cm}$. We want to find the adjacent side RT.

$$\cos(\angle PRT) = \frac{RT}{PR}$$ $$\cos(60^\circ) = \frac{RT}{6}$$

We know $\cos(60^\circ) = \frac{1}{2}$.

$$\frac{1}{2} = \frac{RT}{6}$$ $$RT = 6 \times \frac{1}{2}$$ $$RT = \boxed{\text{3 cm}}$$