Construct triangle PQR with side lengths PQ=6cm, BC=6cm, AC=10cm, and calculate its angles.

The question asks to construct $\triangle PQR$ but provides side lengths $PQ$, $BC$, and $AC$. Assuming $BC$ refers to $QR$ and $AC$ refers to $PR$ for $\triangle PQR$, the side lengths are:
$PQ = 6 \text{ cm}$
$QR = 6 \text{ cm}$
$PR = 10 \text{ cm}$

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

• $PQ + QR > PR \Rightarrow 6 + 6 = 12 \text{ cm}$. Since $12 > 10$, this condition is met.
• $PQ + PR > QR \Rightarrow 6 + 10 = 16 \text{ cm}$. Since $16 > 6$, this condition is met.
• $QR + PR > PQ \Rightarrow 6 + 10 = 16 \text{ cm}$. Since $16 > 6$, this condition is met. Since all conditions are met, the triangle can be constructed.

1. Draw a line segment $PR$ of length $10 \text{ cm}$.
2. With $P$ as the center, use a compass to draw an arc with a radius of $6 \text{ cm}$ (the length of $PQ$).
3. With $R$ as the center, use a compass to draw an arc with a radius of $6 \text{ cm}$ (the length of $QR$).
4. The point where these two arcs intersect is point $Q$.
5. Draw line segments $PQ$ and $QR$ to complete $\triangle PQR$.

Let $p = QR = 6 \text{ cm}$, $q = PR = 10 \text{ cm}$, and $r = PQ = 6 \text{ cm}$.

To find $\angle P$:

To find $\angle R$:
Since $PQ = QR = 6 \text{ cm}$, $\triangle PQR$ is an isosceles triangle, which means $\angle P = \angle R$.

To find $\angle Q$:
The sum of angles in a triangle is $180^\circ$.

The sum of the angles is $33.56^\circ + 33.56^\circ + 112.88^\circ = 180.00^\circ$.
Note: The diagram on the page shows an angle of $60^\circ$ at R, but this contradicts the side lengths given in question 1, so it is not used for this problem.