Trigonometry III

Trigonometry – Sine and Cosine Rules

Lesson Objectives

Understand and apply the sine rule for any triangle.
Understand and apply the cosine rule for any triangle.
Solve problems involving non-right-angled triangles.

Lesson Introduction

When dealing with non-right-angled triangles, we use the Sine Rule and the Cosine Rule to find unknown sides or angles.

Sine Rule

The sine rule states:

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

Used when: we know either two angles and one side (AAS/ASA), or two sides and a non-included angle (SSA).

Cosine Rule

The cosine rule states:

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

Used when: we know two sides and the included angle (SAS), or all three sides (SSS).

Worked Example

Sine Rule Examples

Example 1:
In triangle ABC, \( A = 50^\circ, B = 60^\circ, a = 7 \). Find side \( b \).
\( \frac{7}{\sin(50^\circ)} = \frac{b}{\sin(60^\circ)} \Rightarrow b = \frac{7 \cdot \sin(60^\circ)}{\sin(50^\circ)} \approx 7.84 \)

Example 2:
Given \( a = 10, A = 30^\circ, B = 45^\circ \), find \( b \).
\( b = \frac{10 \cdot \sin(45^\circ)}{\sin(30^\circ)} \approx 14.14 \)

Example 3:
Find angle \( C \) if \( a = 6, c = 10, A = 40^\circ \).
\( \frac{6}{\sin(40^\circ)} = \frac{10}{\sin(C)} \Rightarrow \sin(C) = \frac{10 \cdot \sin(40^\circ)}{6} \approx 1.071 \Rightarrow \text{Not possible (no triangle)} \)

Example 4:
Given triangle DEF: \( D = 70^\circ, F = 60^\circ, d = 12 \). Find side \( f \).
\( E = 180^\circ - 70^\circ - 60^\circ = 50^\circ \)
\( \frac{12}{\sin(70^\circ)} = \frac{f}{\sin(60^\circ)} \Rightarrow f \approx 11.3 \)

Example 5:
\( a = 15, A = 65^\circ, B = 80^\circ \). Find side \( b \).
\( b = \frac{15 \cdot \sin(80^\circ)}{\sin(65^\circ)} \approx 16.4 \)

Cosine Rule Examples

Example 1:
Given \( a = 5, b = 7, C = 60^\circ \). Find side \( c \).
\( c^2 = 5^2 + 7^2 - 2(5)(7)\cos(60^\circ) = 25 + 49 - 70(0.5) = 39 \Rightarrow c = \sqrt{39} \approx 6.24 \)

Example 2:
In triangle PQR, \( p = 10, q = 14, R = 90^\circ \). Find side \( r \).
\( r^2 = 10^2 + 14^2 - 2(10)(14)\cos(90^\circ) = 100 + 196 = 296 \Rightarrow r \approx 17.2 \)

Example 3:
Find angle \( C \) if \( a = 9, b = 8, c = 10 \).
\( \cos C = \frac{9^2 + 8^2 - 10^2}{2(9)(8)} = \frac{145 - 100}{144} = \frac{45}{144} \approx 0.3125 \)
\( C \approx \cos^{-1}(0.3125) \approx 71.8^\circ \)

Example 4:
Triangle XYZ has sides \( x = 11, y = 14, z = 13 \). Find angle \( Z \).
\( \cos Z = \frac{11^2 + 14^2 - 13^2}{2(11)(14)} \approx 0.378 \Rightarrow Z \approx 67.8^\circ \)

Example 5:
[WAEC] Find angle \( B \) if \( a = 12, b = 9, c = 10 \). (Past Question)
\( \cos B = \frac{12^2 + 10^2 - 9^2}{2(12)(10)} = \frac{144 + 100 - 81}{240} = \frac{163}{240} \approx 0.679 \Rightarrow B \approx 47.3^\circ \)

Exercises

Given \( a = 5, A = 50^\circ, B = 60^\circ \), find side b.
In triangle XYZ, \( x = 10, y = 12, X = 30^\circ \). Find angle Y.
[NECO] Triangle has \( a = 11, b = 13, C = 60^\circ \). Find c. (Past Question)
Given sides 7, 8, and 9, find the angle opposite the side of length 9.
Find side c if \( a = 9, b = 10, C = 45^\circ \).
[WAEC] Triangle has \( A = 40^\circ, a = 8, B = 60^\circ \). Find b. (Past Question)
In triangle PQR, \( P = 55^\circ, Q = 65^\circ, p = 9 \). Find side q.
Given \( a = 7, b = 10, c = 11 \), find angle C.
[JAMB] Solve for angle C if \( a = 6, b = 6, c = 6 \). (Past Question)
Given \( a = 9, b = 12, C = 100^\circ \), find side c.

Conclusion/Recap

The sine and cosine rules are powerful tools for solving problems involving non-right-angled triangles. Use the sine rule when dealing with angle-side ratios, and the cosine rule when you have side-angle-side or all three sides.