📐 Level 2 - Topic 4: Law of Sines - Solving Oblique Triangles 🚀△

1) Beyond Right Triangles - Introducing Oblique Triangles and the Law of Sines 🚀

In Level 1, we focused heavily on right-angled triangles and the basic trigonometric ratios (SOH CAH TOA). However, triangles in the real world are not always right-angled. Triangles that do not contain a right angle are called oblique triangles. To solve these triangles (i.e., find unknown sides and angles), we need tools beyond basic trigonometric ratios. Enter the Law of Sines!

Why the Law of Sines?

  • Solving Oblique Triangles: The Law of Sines provides a method to solve triangles when we don't have a right angle and cannot directly use SOH CAH TOA.
  • Real-World Applications: Many practical problems involve oblique triangles, for example, in surveying, navigation, and engineering.
  • Foundation for Further Trigonometry: Understanding the Law of Sines is fundamental for exploring more advanced trigonometric concepts and applications.

In this topic, we will delve into the Law of Sines, understand its formula, learn when and how to apply it, and explore the important concept of the ambiguous case (SSA) where there might be zero, one, or two possible triangles.

2) The Law of Sines - Ratios of Sides and Sines of Opposite Angles ⚖️

The Law of Sines establishes a relationship between the sides of a triangle and the sines of their opposite angles. Consider a triangle \( ABC \) with angles \( A, B, C \) and opposite sides \( a, b, c \) respectively (standard notation: angle at vertex A is \( A \), side opposite to A is \( a \), and so on).

Definition: Law of Sines

In any triangle \( ABC \), the ratio of the length of a side to the sine of its opposite angle is constant for all three sides and angles. Mathematically, this is expressed as:

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

This can also be written in the reciprocal form:

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

Both forms are equivalent and used based on convenience for solving.

Understanding the Formula:

  • The Law of Sines involves pairs of side-angle opposites: side \( a \) is opposite to angle \( A \), side \( b \) is opposite to angle \( B \), and side \( c \) is opposite to angle \( C \).
  • The ratios \( \frac{a}{\sin(A)}, \frac{b}{\sin(B)}, \frac{c}{\sin(C)} \) are all equal to the same constant value for a given triangle. This constant is actually related to the diameter of the circumcircle of the triangle, but we won't delve into that here.
  • To solve for unknowns using the Law of Sines, we typically use two ratios at a time. For example, to solve for side \( a \), if we know angles \( A, B \) and side \( b \), we can use \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} \) to find \( a = \frac{b \sin(A)}{\sin(B)} \).

3) When to Use the Law of Sines - Identifying Applicable Cases 🔍

The Law of Sines is particularly useful in the following cases when solving triangles, given certain information:

  • Case 1: Angle-Side-Angle (ASA) - We know two angles and the included side (the side between the two angles). For example, we know angles \( A, C \) and side \( b \).
  • Case 2: Angle-Angle-Side (AAS) - We know two angles and a non-included side (a side not between the two angles). For example, we know angles \( A, B \) and side \( a \).
  • Case 3: Side-Side-Angle (SSA) - We know two sides and an angle opposite to one of these sides. For example, we know sides \( a, b \) and angle \( A \). This case is known as the ambiguous case, as it might lead to zero, one, or two possible triangles.

Cases Where Law of Sines is NOT Directly Used:

The Law of Sines is generally not directly used in:

  • Side-Angle-Side (SAS) - We know two sides and the included angle. For this, we'll use the Law of Cosines (next topic).
  • Side-Side-Side (SSS) - We know all three sides. Again, we'll use the Law of Cosines for this case.

4) Solving Triangles - Examples Using the Law of Sines 🚀

Example 1: ASA Case

In triangle \( ABC \), \( \angle A = 30^\circ \), \( \angle C = 105^\circ \), and side \( b = 15 \) cm. Solve for the remaining angles and sides.

  1. Find angle \( B \): Sum of angles in a triangle is \( 180^\circ \). \( B = 180^\circ - A - C = 180^\circ - 30^\circ - 105^\circ = 45^\circ \).
  2. Find side \( a \) using Law of Sines: Use \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} \). \( a = \frac{b \sin(A)}{\sin(B)} = \frac{15 \times \sin(30^\circ)}{\sin(45^\circ)} = \frac{15 \times 0.5}{\sqrt{2}/2} = \frac{7.5}{\sqrt{2}/2} = \frac{15}{\sqrt{2}} = \frac{15\sqrt{2}}{2} \approx 10.61 \) cm.
  3. Find side \( c \) using Law of Sines: Use \( \frac{c}{\sin(C)} = \frac{b}{\sin(B)} \). \( c = \frac{b \sin(C)}{\sin(B)} = \frac{15 \times \sin(105^\circ)}{\sin(45^\circ)} \). We can use \( \sin(105^\circ) = \sin(60^\circ + 45^\circ) = \sin(60^\circ)\cos(45^\circ) + \cos(60^\circ)\sin(45^\circ) = \frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2} + \frac{1}{2}\frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} \). So, \( c = \frac{15 \times (\frac{\sqrt{6} + \sqrt{2}}{4})}{\sqrt{2}/2} = \frac{15(\sqrt{6} + \sqrt{2})}{2\sqrt{2}} = \frac{15(\sqrt{3} + 1)}{2} \approx 20.49 \) cm.

Solution: \( \angle B = 45^\circ \), \( a \approx 10.61 \) cm, \( c \approx 20.49 \) cm.

Example 2: AAS Case

In triangle \( ABC \), \( \angle A = 110^\circ \), \( \angle B = 25^\circ \), and side \( a = 25 \) inches. Solve for the remaining angles and sides.

  1. Find angle \( C \): \( C = 180^\circ - A - B = 180^\circ - 110^\circ - 25^\circ = 45^\circ \).
  2. Find side \( b \) using Law of Sines: Use \( \frac{b}{\sin(B)} = \frac{a}{\sin(A)} \). \( b = \frac{a \sin(B)}{\sin(A)} = \frac{25 \times \sin(25^\circ)}{\sin(110^\circ)} \approx \frac{25 \times 0.4226}{0.9397} \approx 11.24 \) inches.
  3. Find side \( c \) using Law of Sines: Use \( \frac{c}{\sin(C)} = \frac{a}{\sin(A)} \). \( c = \frac{a \sin(C)}{\sin(A)} = \frac{25 \times \sin(45^\circ)}{\sin(110^\circ)} \approx \frac{25 \times 0.7071}{0.9397} \approx 18.81 \) inches.

Solution: \( \angle C = 45^\circ \), \( b \approx 11.24 \) inches, \( c \approx 18.81 \) inches.

Example 3: SSA - The Ambiguous Case (One Solution)

In triangle \( ABC \), \( a = 20 \) units, \( b = 15 \) units, and \( \angle A = 40^\circ \). Solve for the remaining angles and sides.

  1. Find \( \sin(B) \) using Law of Sines: Use \( \frac{\sin(B)}{b} = \frac{\sin(A)}{a} \). \( \sin(B) = \frac{b \sin(A)}{a} = \frac{15 \times \sin(40^\circ)}{20} \approx \frac{15 \times 0.6428}{20} \approx 0.4821 \).
  2. Find angle \( B \): \( B = \arcsin(0.4821) \approx 28.82^\circ \). Since \( \sin(B) \) is positive, angle \( B \) could be in the first or second quadrant. However, since \( A = 40^\circ < 90^\circ \) and \( a > b \) (side opposite to \( A \) is longer than side opposite to \( B \)), there is only one possible acute angle for \( B \). So, \( B \approx 28.82^\circ \).
  3. Find angle \( C \): \( C = 180^\circ - A - B \approx 180^\circ - 40^\circ - 28.82^\circ = 111.18^\circ \).
  4. Find side \( c \) using Law of Sines: Use \( \frac{c}{\sin(C)} = \frac{a}{\sin(A)} \). \( c = \frac{a \sin(C)}{\sin(A)} = \frac{20 \times \sin(111.18^\circ)}{\sin(40^\circ)} \approx \frac{20 \times 0.9325}{0.6428} \approx 29.03 \) units.

Solution: \( \angle B \approx 28.82^\circ \), \( \angle C \approx 111.18^\circ \), \( c \approx 29.03 \) units.

Example 4: SSA - The Ambiguous Case (Two Solutions)

In triangle \( ABC \), \( a = 10 \) units, \( b = 12 \) units, and \( \angle A = 30^\circ \). Solve for possible triangles.

  1. Find \( \sin(B) \) using Law of Sines: Use \( \frac{\sin(B)}{b} = \frac{\sin(A)}{a} \). \( \sin(B) = \frac{b \sin(A)}{a} = \frac{12 \times \sin(30^\circ)}{10} = \frac{12 \times 0.5}{10} = 0.6 \).
  2. Find angle \( B \): \( B_1 = \arcsin(0.6) \approx 36.87^\circ \). Since \( \sin(B) \) is positive, there could be a second angle in the second quadrant: \( B_2 = 180^\circ - B_1 \approx 180^\circ - 36.87^\circ = 143.13^\circ \). We need to check if both are valid.
  3. Check for Triangle 1 (using \( B_1 \approx 36.87^\circ \)): \( A + B_1 = 30^\circ + 36.87^\circ = 66.87^\circ < 180^\circ \). Valid. \( C_1 = 180^\circ - A - B_1 \approx 180^\circ - 30^\circ - 36.87^\circ = 113.13^\circ \). Find \( c_1 \) using Law of Sines: \( c_1 = \frac{a \sin(C_1)}{\sin(A)} = \frac{10 \times \sin(113.13^\circ)}{\sin(30^\circ)} \approx \frac{10 \times 0.9196}{0.5} \approx 18.39 \) units.
  4. Check for Triangle 2 (using \( B_2 \approx 143.13^\circ \)): \( A + B_2 = 30^\circ + 143.13^\circ = 173.13^\circ < 180^\circ \). Valid. \( C_2 = 180^\circ - A - B_2 \approx 180^\circ - 30^\circ - 143.13^\circ = 6.87^\circ \). Find \( c_2 \) using Law of Sines: \( c_2 = \frac{a \sin(C_2)}{\sin(A)} = \frac{10 \times \sin(6.87^\circ)}{\sin(30^\circ)} \approx \frac{10 \times 0.1195}{0.5} \approx 2.39 \) units.

Solution: Two possible triangles exist:

Triangle 1: \( \angle B_1 \approx 36.87^\circ \), \( \angle C_1 \approx 113.13^\circ \), \( c_1 \approx 18.39 \) units.

Triangle 2: \( \angle B_2 \approx 143.13^\circ \), \( \angle C_2 \approx 6.87^\circ \), \( c_2 \approx 2.39 \) units.

The Ambiguous Case (SSA) - Summary:

When using the Law of Sines with SSA case, always check for the possibility of two solutions. After finding a possible angle \( B_1 = \arcsin(\frac{b \sin(A)}{a}) \), calculate the potential second angle \( B_2 = 180^\circ - B_1 \). Then, verify if \( A + B_2 < 180^\circ \).

  • If \( \sin(B) > 1 \): No solution (no triangle exists).
  • If \( \sin(B) = 1 \): One right-angled solution.
  • If \( \sin(B) < 1 \) and \( a \geq b \): One solution (acute angle \( B \)).
  • If \( \sin(B) < 1 \) and \( a < b \): Possible two solutions (acute \( B_1 \) and obtuse \( B_2 = 180^\circ - B_1 \)), check if both are valid by ensuring \( A + B_2 < 180^\circ \). It's also possible to have one or zero solutions in this case based on the height of the triangle relative to side \( a \).

5) Practice Questions 🎯

5.1 Fundamental – Applying Law of Sines (ASA, AAS Cases)

1. In triangle \( PQR \), \( \angle P = 42^\circ \), \( \angle Q = 75^\circ \), and side \( r = 12 \) cm. Find side \( p \).

2. In triangle \( XYZ \), \( \angle X = 120^\circ \), \( \angle Y = 35^\circ \), and side \( y = 20 \) inches. Find side \( x \).

3. In triangle \( ABC \), \( \angle A = 50^\circ \), \( \angle B = 60^\circ \), and side \( c = 15 \) units. Find side \( a \) and side \( b \).

4. Given triangle \( DEF \) with \( \angle D = 25^\circ \), \( \angle E = 95^\circ \), and side \( f = 8 \) meters. Find side \( d \) and side \( e \).

5. In triangle \( KLM \), \( \angle K = 30^\circ \), \( \angle M = 110^\circ \), and side \( l = 24 \) cm. Find angle \( L \), side \( k \) and side \( m \).

5.2 Challenging – SSA Ambiguous Case and Problem Solving 💪🚀

1. In triangle \( ABC \), \( a = 25 \) units, \( b = 30 \) units, and \( \angle A = 50^\circ \). Solve for all possible triangles.

2. In triangle \( XYZ \), \( x = 15 \) cm, \( y = 10 \) cm, and \( \angle Y = 65^\circ \). Solve for all possible triangles.

3. Triangle \( ABC \) has \( a = 8 \), \( \angle A = 60^\circ \), and \( b = 9 \). Determine if there is one, two, or no possible triangles. If there are solutions, find them.

4. A surveyor needs to measure the distance across a river. From point A on one bank, they sight a point C on the opposite bank at an angle of \( 28^\circ \). Then they move 200 meters along the bank to point B and sight point C at an angle of \( 48^\circ \). Assuming the riverbank is straight, find the distance across the river (from point A perpendicular to the line of sight AC to the riverbank at C).

5. (Conceptual) Explain geometrically why the SSA case can be ambiguous. Use diagrams to illustrate the possibilities of zero, one, or two triangles when given two sides and a non-included angle.

6) Summary - Mastering the Law of Sines for Oblique Triangles 🎉

  • Law of Sines: \( \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \) - Relates sides of a triangle to sines of opposite angles.
  • Solving Oblique Triangles: Used to solve triangles that are not right-angled.
  • Applicable Cases: ASA, AAS, SSA (Ambiguous Case).
  • Ambiguous Case (SSA): Can result in zero, one, or two possible triangles. Always check for potential multiple solutions.
  • Steps to Solve: Identify the known case (ASA, AAS, SSA), set up Law of Sines ratios, solve for unknowns, check for ambiguous case when SSA.
  • Applications: Useful in various fields involving triangle measurements where right angles are not guaranteed.

Congratulations! You've now learned the Law of Sines, a powerful tool for solving oblique triangles! Understanding when and how to apply it, especially being aware of the ambiguous SSA case, will significantly enhance your trigonometry problem-solving skills. In the next topic, we'll explore another essential tool for solving triangles - the Law of Cosines, which is particularly useful in cases where the Law of Sines is not directly applicable. Keep practicing and you'll become proficient in solving any triangle! 📐🚀△🌟

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