Complete collection of examples for Law of Sines, Law of Cosines, and Area Formulas
← Back to CourseProblem: In triangle ABC, a = 10, A = 30°, B = 45°. Find side b.
Solution:
Using Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B}$$
$$b = a \times \frac{\sin B}{\sin A} = 10 \times \frac{\sin 45°}{\sin 30°}$$
$$b = 10 \times \frac{0.7071}{0.5} = 10 \times 1.4142 = 14.14$$
Answer: b ≈ 14.14
Problem: In triangle ABC, a = 8, b = 6, A = 50°. Find angle B.
Solution:
$$\sin B = \frac{b \sin A}{a} = \frac{6 \times \sin 50°}{8} = \frac{6 \times 0.766}{8} = 0.5745$$
$$B = \arcsin(0.5745) = 35.1°$$
Note: Ambiguous case - check if B = 180° - 35.1° = 144.9° is also valid. Since A + B = 50° + 144.9° = 194.9° > 180°, only B = 35.1° works.
Answer: B ≈ 35.1°
Problem: In triangle ABC, a = 10, b = 12, A = 40°. How many triangles are possible?
Solution:
$$\sin B = \frac{b \sin A}{a} = \frac{12 \times \sin 40°}{10} = \frac{12 \times 0.6428}{10} = 0.7714$$
$$B_1 = \arcsin(0.7714) = 50.5°$$
$$B_2 = 180° - 50.5° = 129.5°$$
Check: A + B₁ = 40° + 50.5° = 90.5° < 180° ✓
Check: A + B₂ = 40° + 129.5° = 169.5° < 180° ✓
Answer: Two triangles are possible with B = 50.5° or B = 129.5°
Problem: In triangle ABC, a = 7, A = 60°. Find the circumradius R.
Solution:
$$R = \frac{a}{2\sin A} = \frac{7}{2 \times \sin 60°} = \frac{7}{2 \times 0.866} = \frac{7}{1.732} = 4.04$$
Answer: R ≈ 4.04
Problem: In triangle ABC, a = 5, b = 7, C = 60°. Find side c.
Solution:
$$c^2 = a^2 + b^2 - 2ab\cos C$$
$$c^2 = 5^2 + 7^2 - 2(5)(7)\cos 60°$$
$$c^2 = 25 + 49 - 70 \times 0.5 = 74 - 35 = 39$$
$$c = \sqrt{39} = 6.24$$
Answer: c ≈ 6.24
Problem: In triangle ABC, a = 5, b = 6, c = 7. Find angle C.
Solution:
$$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{25 + 36 - 49}{2(5)(6)} = \frac{12}{60} = 0.2$$
$$C = \arccos(0.2) = 78.46°$$
Answer: C ≈ 78.46°
Problem: A triangle has sides 8, 10, and 13. Is it acute, right, or obtuse?
Solution:
Compare c² with a² + b² (where c = 13 is the longest side):
$$c^2 = 13^2 = 169$$
$$a^2 + b^2 = 8^2 + 10^2 = 64 + 100 = 164$$
Since 169 > 164, we have c² > a² + b², so the triangle is obtuse.
Answer: Obtuse triangle (the angle opposite the longest side is greater than 90°)
Problem: A ship sails 10 km east, then turns 120° and sails 8 km. How far is it from the starting point?
Solution:
The angle between the two paths is 180° - 120° = 60° (interior angle).
$$d^2 = 10^2 + 8^2 - 2(10)(8)\cos 60°$$
$$d^2 = 100 + 64 - 160 \times 0.5 = 164 - 80 = 84$$
$$d = \sqrt{84} = 9.17 \text{ km}$$
Answer: Approximately 9.17 km from the starting point
Problem: A triangle has base 10 cm and height 6 cm. Find its area.
Solution:
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 6 = 30 \text{ cm}^2$$
Answer: 30 cm²
Problem: In triangle ABC, a = 8, b = 5, C = 30°. Find the area.
Solution:
$$\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2} \times 8 \times 5 \times \sin 30°$$
$$= \frac{1}{2} \times 8 \times 5 \times 0.5 = 10$$
Answer: 10 square units
Problem: Find the area of a triangle with sides 5, 6, 7.
Solution:
Step 1: Find semi-perimeter: $$s = \frac{5+6+7}{2} = 9$$
Step 2: Apply Heron's formula:
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{9 \times 4 \times 3 \times 2}$$
$$= \sqrt{216} = 6\sqrt{6} \approx 14.70$$
Answer: ≈ 14.70 square units
Problem: A triangle has sides 13, 14, 15. Find the height to the side of length 14.
Solution:
Step 1: Find area using Heron's formula: s = (13+14+15)/2 = 21
$$\text{Area} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$$
Step 2: Use Area = ½bh to find height:
$$84 = \frac{1}{2} \times 14 \times h$$
$$h = \frac{84 \times 2}{14} = 12$$
Answer: Height = 12 units
Problem: Find the area of an equilateral triangle with side 8.
Solution:
For an equilateral triangle: $$\text{Area} = \frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4} \times 64 = 16\sqrt{3} \approx 27.71$$
Answer: 16√3 ≈ 27.71 square units