To solve the triangle $\triangle PQR$ using the Law of Sines, we will follow these steps:

Given Information:

• $\angle P = 99^\circ$
• $\angle Q = 31^\circ$
• $PQ = 11$ (side opposite $\angle R$)

Step 1: Find $\angle R$

Since the sum of angles in a triangle is always $180^\circ$, we can calculate $\angle R$ as:

$\angle R = 180^\circ - \angle P - \angle Q = 180^\circ - 99^\circ - 31^\circ = 50^\circ$

Thus, $\angle R = 50^\circ$.

Step 2: Apply the Law of Sines

The Law of Sines states that: $\frac{PR}{\sin(\angle Q)} = \frac{PQ}{\sin(\angle R)} = \frac{QR}{\sin(\angle P)}$

Using this, we will calculate the lengths of sides $PR$ and $QR$.

Solve for $PR$

We know $PQ = 11$, $\angle R = 50^\circ$, and $\angle Q = 31^\circ$, so:

$\frac{PR}{\sin(31^\circ)} = \frac{11}{\sin(50^\circ)}$

Solving for $PR$:

$PR = \frac{11 \cdot \sin(31^\circ)}{\sin(50^\circ)}$

Solve for $QR$

Similarly, using $PQ = 11$ and $\angle P = 99^\circ$:

$\frac{QR}{\sin(99^\circ)} = \frac{11}{\sin(50^\circ)}$

Solving for $QR$:

$QR = \frac{11 \cdot \sin(99^\circ)}{\sin(50^\circ)}$

Let me compute these values.### Solution:

• $\angle R = 50^\circ$
• $PR \approx 7.40$
• $QR \approx 14.18$

These values have been rounded to the nearest hundredth, as requested.

Would you like further clarification or details on any step? Here are some related questions for practice:

1. How would the Law of Sines change if we were given different sides or angles?
2. Can you apply the Law of Cosines to verify the side lengths?
3. What is the significance of the Law of Sines in non-right triangles?
4. How do you calculate the area of this triangle using the sine rule?
5. How can you verify the calculated angles sum to $180^\circ$?

Tip: Always check that the sum of the angles in any triangle equals $180^\circ$, a useful verification step.