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The problem requires calculating the third side of a triangle using the law of cosines:

\[
c^2 = a^2 + b^2 - 2ab\cos(\gamma)
\]

Let's solve both cases.

---

### **Case (a)**:
Given:
\[
a = 4 \, \text{cm}, \, b = 5 \, \text{cm}, \, \gamma = 60^\circ
\]
Using the formula:
\[
c^2 = 4^2 + 5^2 - 2 \cdot 4 \cdot 5 \cdot \cos(60^\circ)
\]
\[
c^2 = 16 + 25 - 40 \cdot \frac{1}{2}
\]
\[
c^2 = 41 - 20 = 21
\]
\[
c = \sqrt{21} \approx 4.58 \, \text{cm}
\]

---

### **Case (b)**:
Given:
\[
a = 6 \, \text{cm}, \, b = 9 \, \text{cm}, \, \gamma = 135^\circ
\]
Since \(\cos(135^\circ) = -\cos(45^\circ) = -\frac{\sqrt{2}}{2}\):
\[
c^2 = 6^2 + 9^2 - 2 \cdot 6 \cdot 9 \cdot \left(-\frac{\sqrt{2}}{2}\right)
\]
\[
c^2 = 36 + 81 + 54\sqrt{2}
\]
\[
c^2 \approx 117 + 76.37 = 193.37
\]
\[
c \approx \sqrt{193.37} \approx 13.91 \, \text{cm}
\]

---

### **Final Results**:
1. For case (a): \( c \approx 4.58 \, \text{cm} \)
2. For case (b): \( c \approx 13.91 \, \text{cm} \)

Would you like a detailed explanation of any step or further assistance?

---

### Related Questions:
1. How does the law of cosines relate to the Pythagorean theorem?
2. How would the results change if the angle \(\gamma\) were \(90^\circ\)?
3. What are other ways to calculate the third side of a triangle?
4. How do we simplify calculations involving square roots?
5. What happens if the given angle \(\gamma\) is obtuse or acute?

---

### Tip:
Always ensure that angles are in the correct units (degrees or radians) when applying trigonometric functions.

Dwa boki trójkąta mają długość a i b, a kąt trójkąta zawarty między tymi bokami jest równy γ. Oblicz długość trzeciego boku tego trójkąta, jeśli a) a = 4 cm, b = 5 cm, γ = 60°, b) a = 6 cm, b = 9 cm, γ = 135°.

Learn how to calculate the third side of a triangle given two sides and the included angle using the law of cosines. This problem demonstrates detailed solutions for cases with angles of 60° and 135°. Perfect for Grades 10-12 geometry and trigonometry practice.

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