Solved problems: general triangles using Law of Sines and Cosines

📚 Introduction to Solving General Triangles

General triangles, unlike right triangles, don't have a 90-degree angle. Solving them means finding all unknown sides and angles using given information. The Law of Sines and the Law of Cosines are the key tools to achieve this.

📜 A Brief History

The principles behind the Law of Sines and Cosines have ancient roots, with early concepts appearing in Greek trigonometry. Islamic mathematicians further developed these ideas, and they were later refined and popularized in Europe during the Renaissance. These laws allow us to perform calculations that were impossible using only right-triangle trigonometry.

📐 Key Principles: Law of Sines

The Law of Sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle. Mathematically, it's expressed as:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

• 🔍 When to use: When you know two angles and one side (AAS or ASA) or two sides and an angle opposite one of them (SSA). The SSA case can be ambiguous, potentially leading to zero, one, or two possible triangles.
• 💡 Example: If you know angles A and B and side a, you can find side b using the formula: $b = \frac{a \sin B}{\sin A}$.
• 📝 Ambiguous Case (SSA): Be cautious! Draw a diagram to visualize possible solutions. Check if the side opposite the known angle is long enough to form a triangle.

📏 Key Principles: Law of Cosines

The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. There are three common formulas:

$a^2 = b^2 + c^2 - 2bc \cos A$

$b^2 = a^2 + c^2 - 2ac \cos B$

$c^2 = a^2 + b^2 - 2ab \cos C$

• 🧭 When to use: When you know three sides (SSS) or two sides and the included angle (SAS).
• 🧮 Example (SSS): If you know sides a, b, and c, you can find angle A using the formula: $A = \cos^{-1}(\frac{b^2 + c^2 - a^2}{2bc})$.
• ✨ Example (SAS): If you know sides b and c, and angle A, you can find side a using the formula: $a = \sqrt{b^2 + c^2 - 2bc \cos A}$.

✍️ Worked Examples

Let's tackle some common scenarios:

Example 1: AAS

Given: $A = 30^\circ$, $B = 70^\circ$, and $a = 8 \text{ cm}$. Find side $b$.

Solution: Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$

$b = \frac{a \sin B}{\sin A} = \frac{8 \sin 70^\circ}{\sin 30^\circ} \approx 15.04 \text{ cm}$

Example 2: SAS

Given: $b = 5 \text{ cm}$, $c = 8 \text{ cm}$, and $A = 77^\circ$. Find side $a$.

Solution: Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$

$a = \sqrt{5^2 + 8^2 - 2(5)(8) \cos 77^\circ} \approx 8.56 \text{ cm}$

Example 3: SSS

Given: $a = 7 \text{ cm}$, $b = 9 \text{ cm}$, and $c = 5 \text{ cm}$. Find angle $A$.

Solution: Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$

$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{9^2 + 5^2 - 7^2}{2(9)(5)} = \frac{57}{90}$

$A = \cos^{-1}(\frac{57}{90}) \approx 50.70^\circ$

✍️ Practice Quiz

Solve the following triangle problems:

1. ❓ Given $A = 45^\circ$, $B = 60^\circ$, and $a = 10$, find $b$.
2. 📐 Given $b = 7$, $c = 10$, and $A = 30^\circ$, find $a$.
3. 📏 Given $a = 5$, $b = 8$, and $c = 11$, find $A$.
4. 🌍 Given $B = 100^\circ$, $C = 50^\circ$, and $b = 25$, find $c$.
5. 💡 Given $a = 12$, $c = 15$, and $B = 54^\circ$, find $b$.
6. 🧪 Given $a = 6$, $b = 8$, and $c = 10$, find $C$.
7. 🧬 Given $A = 110^\circ$, $b = 15$, and $c = 10$, find $a$.

🔑 Conclusion

Mastering the Law of Sines and the Law of Cosines unlocks the ability to solve any triangle. Understanding when to apply each law is crucial. Remember to consider the ambiguous case (SSA) carefully. Practice is key! Good luck!