Construct a triangle PQR with sides PQ = 8 cm, QR = 11 cm, and PR = 10 cm. Verify the triangle inequality and calculate its angles using the Law of Cosines.

We apply the Triangle Inequality Theorem: the sum of any two sides must be greater than the third side.

• $PQ + QR > PR \Rightarrow 8 + 11 = 19 \text{ cm}$. Since $19 > 10$, this condition is met.
• $PQ + PR > QR \Rightarrow 8 + 10 = 18 \text{ cm}$. Since $18 > 11$, this condition is met.
• $QR + PR > PQ \Rightarrow 11 + 10 = 21 \text{ cm}$. Since $21 > 8$, this condition is met. Since all conditions are met, the triangle can be constructed.

1. Draw a line segment $PQ$ of length $8 \text{ cm}$.
2. With $P$ as the center, use a compass to draw an arc with a radius of $10 \text{ cm}$ (the length of $PR$).
3. With $Q$ as the center, use a compass to draw an arc with a radius of $11 \text{ cm}$ (the length of $QR$).
4. The point where these two arcs intersect is point $R$.
5. Draw line segments $PR$ and $QR$ to complete $\triangle PQR$.

Let $p = QR = 11 \text{ cm}$, $q = PR = 10 \text{ cm}$, and $r = PQ = 8 \text{ cm}$.

To find $\angle P$:

To find $\angle Q$:

To find $\angle R$:

The sum of the angles is $74.38^\circ + 61.05^\circ + 44.57^\circ = 180.00^\circ$.