Law of Sines and Cosines
Solving oblique triangles with side and angle relationships
Law of Sines and Cosines
Solving oblique triangles with side and angle relationships
Math - Grade 9-12
- 1
In triangle ABC, angle A = 35 degrees, angle B = 65 degrees, and side a = 12. Find side b.
Match each side with its opposite angle before substituting into the formula.
Angle C = 80 degrees. Using the Law of Sines, b / sin(65 degrees) = 12 / sin(35 degrees), so b = 12 sin(65 degrees) / sin(35 degrees) about 19.0. - 2
In triangle ABC, angle A = 42 degrees, side a = 9, and side b = 13. Find angle B if it exists.
This is an SSA case, so check for the ambiguous case.
Using the Law of Sines, sin(B) / 13 = sin(42 degrees) / 9, so sin(B) about 0.9666. This gives B about 75.2 degrees or B about 104.8 degrees. Both are possible because each makes the angle sum less than 180 degrees, so there are two solutions. - 3
In triangle ABC, side a = 7, side b = 10, and angle C = 48 degrees. Find side c.
Using the Law of Cosines, c squared = 7 squared + 10 squared - 2(7)(10)cos(48 degrees). This gives c squared about 55.3, so c about 7.4. - 4
In triangle ABC, side a = 15, side b = 18, and side c = 21. Find angle C.
When all three sides are known, solve for the cosine of the angle first.
Using the Law of Cosines, 21 squared = 15 squared + 18 squared - 2(15)(18)cos(C). Solving gives cos(C) about 0.2, so angle C about 78.5 degrees. - 5
A surveyor marks two sides of a triangular field as 120 meters and 150 meters with an included angle of 52 degrees. Find the third side of the field.
Using the Law of Cosines, c squared = 120 squared + 150 squared - 2(120)(150)cos(52 degrees). This gives c squared about 14742.6, so the third side is about 121.4 meters. - 6
In triangle ABC, angle A = 28 degrees, angle C = 91 degrees, and side c = 16. Find side a.
Use the known side and its opposite angle as your starting ratio.
Using the Law of Sines, a / sin(28 degrees) = 16 / sin(91 degrees). So a = 16 sin(28 degrees) / sin(91 degrees) about 7.5. - 7
A triangle has sides 9 and 14 with an included angle of 120 degrees. Find the area of the triangle.
The area formula for two sides and the included angle is Area = 1/2 ab sin(C). So Area = 1/2 (9)(14)sin(120 degrees) about 54.6 square units. - 8
In triangle ABC, side a = 11, side b = 17, and angle A = 32 degrees. Find angle B if it exists.
Check both the acute angle and its supplement.
Using the Law of Sines, sin(B) / 17 = sin(32 degrees) / 11, so sin(B) about 0.8184. This gives B about 54.9 degrees or B about 125.1 degrees. Both are valid because each leads to a positive third angle, so there are two solutions. - 9
In triangle ABC, side a = 8, side b = 13, and side c = 16. Find angle A.
Using the Law of Cosines, 8 squared = 13 squared + 16 squared - 2(13)(16)cos(A). Solving gives cos(A) about 0.8678, so angle A about 29.8 degrees. - 10
Two roads leave a town at an angle of 37 degrees. A point on one road is 18 miles from town, and a point on the other road is 25 miles from town. How far apart are the two points?
The distance between the two points is the side opposite the included angle.
The roads and the line between the two points form a triangle with sides 18 and 25 and included angle 37 degrees. Using the Law of Cosines, c squared = 18 squared + 25 squared - 2(18)(25)cos(37 degrees), so c about 15.6 miles. - 11
In triangle ABC, angle B = 47 degrees, angle C = 68 degrees, and side b = 20. Find side c.
Using the Law of Sines, c / sin(68 degrees) = 20 / sin(47 degrees). So c = 20 sin(68 degrees) / sin(47 degrees) about 25.3. - 12
A triangle has sides 13, 15, and 20. Find its area by first using the Law of Cosines to find one angle.
Find an angle first, then apply the area formula 1/2 ab sin(C).
Use the Law of Cosines with the angle between sides 13 and 15. Since 20 squared = 13 squared + 15 squared - 2(13)(15)cos(C), cos(C) about -0.0154, so C about 90.9 degrees. Then Area = 1/2 (13)(15)sin(90.9 degrees) about 97.5 square units. - 13
In triangle ABC, angle A = 25 degrees, side a = 6, and side b = 14. Determine whether the triangle has zero, one, or two possible solutions.
Using the Law of Sines, sin(B) / 14 = sin(25 degrees) / 6, so sin(B) about 0.9861. This gives B about 80.4 degrees or B about 99.6 degrees. Both are possible because each leaves a positive third angle, so the triangle has two possible solutions. - 14
In triangle ABC, side a = 10, side b = 12, and side c = 5. Find the smallest angle.
Identify the shortest side first to know which angle is smallest.
The smallest angle is opposite the shortest side, so it is angle C. Using the Law of Cosines, 5 squared = 10 squared + 12 squared - 2(10)(12)cos(C). Solving gives cos(C) about 0.9125, so angle C about 24.2 degrees. - 15
A triangular garden has sides 22 feet and 30 feet with an included angle of 75 degrees. Find the third side and then find the area.
This problem uses both formulas, one for a side length and one for area.
First use the Law of Cosines: c squared = 22 squared + 30 squared - 2(22)(30)cos(75 degrees), so c about 30.0 feet. Then use the area formula: Area = 1/2 (22)(30)sin(75 degrees) about 318.7 square feet.