To solve the triangle $PQR$ with the given dataTo solve the triangle $PQR$ with the given

• $QR = 42.2 \, \text{cm}$ (side opposite to $\angle P$)
• $PQ = 21.2 \, \text{cm}$ (side opposite to $\angle R$)
• $\angle P = 100.5^\circ$

We need to determine the remaining angle $\angle Q$ and $\angle R$, as well as the side $PR$.

Step 1: Use the Law of Sines

The Law of Sines states:

$\frac{QR}{\sin P} = \frac{PR}{\sin Q} = \frac{PQ}{\sin R}$

We know $\angle P$, $QR$, and $PQ$. So, we can find the unknown angle $\angle R$ first.

$\frac{QR}{\sin P} = \frac{PQ}{\sin R}$

Step 2: Solve for $\sin R$

$\sin R = \frac{PQ \times \sin P}{QR}$ $\sin R = \frac{21.2 \times \sin(100.5^\circ)}{42.2}$

Calculate:

$\sin(100.5^\circ) \approx 0.9848$ $\sin R = \frac{21.2 \times 0.9848}{42.2} \approx \frac{20.86816}{42.2} \approx 0.4944$

Thus,

$R \approx \arcsin(0.4944) \approx 29.6^\circ$

Step 3: Determine $\angle Q$

Using the angle sum property of a triangle:

$\angle Q = 180^\circ - \angle P - \angle R$ $\angle Q = 180^\circ - 100.5^\circ - 29.6^\circ = 49.9^\circ$

Step 4: Calculate the side $PR$ using the Law of Sines

$\frac{PR}{\sin Q} = \frac{QR}{\sin P}$ $PR = \frac{QR \times \sin Q}{\sin P}$ $PR = \frac{42.2 \times \sin(49.9^\circ)}{\sin(100.5^\circ)}$

Calculate:

$\sin(49.9^\circ) \approx 0.766$ $PR = \frac{42.2 \times 0.766}{0.9848} \approx \frac{32.3352}{0.9848} \approx 32.8 \, \text{cm}$

Summary:

• $\angle R \approx 29.6^\circ$
• $\angle Q \approx 49.9^\circ$
• $PR \approx 32.8 \, \text{cm}$

Would you like more details or have any questions?

Related Questions:

1. How can you verify these results using the Law of Cosines?
2. What would the area of triangle $PQR$ be using Heron's formula?
3. How can you find the altitude from $P$ to side $QR$?
4. What is the inradius of triangle $PQR$?
5. How does the obtuse angle $\angle P$ affect the triangle's properties?

Tip:

In triangles with an obtuse angle, always ensure to use the correct sine values and carefully consider the relationships between the angles and sides.