Construct a triangle P.Q.R such that P.Q=6cm and angle P.Q.R = 70 degrees. ii. Construct perpendicular by setter of PQ and PR and let the inter section of the mediators be O. iii. Draw a circle with O as center and PO as radius.

• Place a ruler along the line segment $RP$ on your constructed triangle.
• Read the length from the ruler.

Based on the given dimensions, the theoretical length of $RP$ can be calculated using the Law of Cosines:

$$RP^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)$$ $$RP^2 = (6 \text{ cm})^2 + (4 \text{ cm})^2 - 2(6 \text{ cm})(4 \text{ cm})\cos(70^\circ)$$ $$RP^2 = 36 \text{ cm}^2 + 16 \text{ cm}^2 - 48 \text{ cm}^2 \times 0.3420$$ $$RP^2 = 52 \text{ cm}^2 - 16.416 \text{ cm}^2$$ $$RP^2 = 35.584 \text{ cm}^2$$ $$RP = \sqrt{35.584 \text{ cm}^2} \approx \boxed{5.97 \text{ cm}}$$

Your measured value should be close to this.

• The angle $\angle PQR$ is given in the problem as $70^\circ$.
• To measure $\angle QPR$: Place the center of your protractor at point $P$ and align the baseline with $PQ$. Read the angle where $PR$ crosses the protractor scale.
• To measure $\angle PRQ$: Place the center of your protractor at point $R$ and align the baseline with $QR$. Read the angle where $PR$ crosses the protractor scale.

Based on the given dimensions, the theoretical values for the other angles are:
Using the Law of Sines:

$$\frac{\sin(\angle QPR)}{QR} = \frac{\sin(\angle PQR)}{RP}$$ $$\sin(\angle QPR) = \frac{4 \text{ cm} \times \sin(70^\circ)}{5.97 \text{ cm}} \approx 0.629$$ $$\angle QPR = \arcsin(0.629) \approx \boxed{39.0^\circ}$$

Then, using the sum of angles in a triangle:

$$\angle PRQ = 180^\circ - \angle PQR - \angle QPR$$ $$\angle PRQ = 180^\circ - 70^\circ - 39.0^\circ = \boxed{71.0^\circ}$$

Your measured values for $\angle QPR$ and $\angle PRQ$ should be close to these.