Using the Law of Sines to Solve Triangle Problems

Problem Statement: In triangle \(ABC\), given \(A = 106^\circ\), \(B = 31^\circ\), and side \(a = 10\) cm. Solve the triangle by finding angle \(C\) and sides \(b\) and \(c\). Round answers to one decimal place.

Solution:

1. Find angle \(C\) using the angle sum of a triangle: \[ C = 180^\circ - (A + B) = 180^\circ - (106^\circ + 31^\circ) = 43^\circ \]
2. Use the Law of Sines to find side \(b\): \[ \frac{a}{\sin A} = \frac{b}{\sin B} \implies b = \frac{a \sin B}{\sin A} = \frac{10 \sin 31^\circ}{\sin 106^\circ} \approx 5.4 \text{ cm} \]
3. Use the Law of Sines to find side \(c\): \[ \frac{a}{\sin A} = \frac{c}{\sin C} \implies c = \frac{a \sin C}{\sin A} = \frac{10 \sin 43^\circ}{\sin 106^\circ} \approx 7.1 \text{ cm} \]

Problem Statement: The angles of elevation to the top \(C\) of a building from two points \(A\) and \(B\) on level ground are \(50^\circ\) and \(60^\circ\), respectively. The distance between \(A\) and \(B\) is 30 meters. Points \(A\), \(B\), and \(C\) are in the same vertical plane. Find the height \(h\) of the building. Round to the nearest meter.

Solution:

1. In triangle \(ABC\), angle \(B = 180^\circ - 60^\circ = 120^\circ\) (supplementary).
2. Angle \(C = 180^\circ - (50^\circ + 120^\circ) = 10^\circ\).
3. Apply the Law of Sines to find \(d\) (distance from \(A\) to building base): \[ \frac{d}{\sin 50^\circ} = \frac{30}{\sin 10^\circ} \implies d = \frac{30 \sin 50^\circ}{\sin 10^\circ} \]
4. In the right triangle: \(\sin 60^\circ = \frac{h}{d} \implies h = d \sin 60^\circ\).
5. Combine and calculate: \[ h = \frac{30 \sin 50^\circ \sin 60^\circ}{\sin 10^\circ} \approx 115 \text{ meters} \]

Problem Statement: In triangle \(ABC\), given side \(a = 12\) cm, side \(b = 19\) cm, and angle \(A = 80^\circ\) (angle \(A\) is opposite side \(a\)). Find side \(c\) and angles \(B\) and \(C\) if possible. Round answers to one decimal place.

Solution:

1. Apply the Law of Sines: \[ \frac{a}{\sin A} = \frac{b}{\sin B} \implies \sin B = \frac{b \sin A}{a} = \frac{19 \sin 80^\circ}{12} \approx 1.56 \]
2. Since \(\sin B > 1\), no real angle \(B\) exists. This triangle has no solution.

Problem Statement: In triangle \(ABC\), given side \(a = 14\) cm, side \(b = 19\) cm, and angle \(A = 32^\circ\) (angle \(A\) is opposite side \(a\)). Find side \(c\) and angles \(B\) and \(C\) if possible. Round answers to one decimal place.

Solution: This is an ambiguous case (SSA) with two possible triangles.

1. Apply the Law of Sines: \[ \frac{a}{\sin A} = \frac{b}{\sin B} \implies \sin B = \frac{b \sin A}{a} = \frac{19 \sin 32^\circ}{14} \approx 0.7192 \]
2. Two angles satisfy \(\sin B = 0.7192\): \[ B_1 = \arcsin(0.7192) \approx 46.0^\circ \quad \text{and} \quad B_2 = 180^\circ - 46.0^\circ = 134.0^\circ \]

Solution 1:

• \(C_1 = 180^\circ - A - B_1 = 180^\circ - 32^\circ - 46.0^\circ = 102.0^\circ\)
• \(c_1 = \frac{a \sin C_1}{\sin A} = \frac{14 \sin 102.0^\circ}{\sin 32^\circ} \approx 25.8 \text{ cm}\)

Solution 2:

• \(C_2 = 180^\circ - A - B_2 = 180^\circ - 32^\circ - 134.0^\circ = 14.0^\circ\)
• \(c_2 = \frac{a \sin C_2}{\sin A} = \frac{14 \sin 14.0^\circ}{\sin 32^\circ} \approx 6.4 \text{ cm}\)

Question: Triangle \(ABC\) has \(A = 104^\circ\), \(C = 33^\circ\), and side \(c = 9\) m. Solve the triangle by finding angle \(B\) and sides \(a\) and \(b\). Round answers to one decimal place.

Solution:

• \(B = 180^\circ - (104^\circ + 33^\circ) = 43^\circ\)
• \(a = \frac{c \sin A}{\sin C} = \frac{9 \sin 104^\circ}{\sin 33^\circ} \approx 16.0 \text{ m}\)
• \(b = \frac{c \sin B}{\sin C} = \frac{9 \sin 43^\circ}{\sin 33^\circ} \approx 11.3 \text{ m}\)

Question: Redo Problem 2 with the distance between points \(A\) and \(B\) equal to 50 meters.

Solution:

Using the same method as Problem 2 with \(AB = 50\) m:

\[ h = \frac{50 \sin 50^\circ \sin 60^\circ}{\sin 10^\circ} \approx 191 \text{ meters} \]