Law of Sines Section 6.1

So far we have learned how to solve for only one type of triangle Right Triangles Next, we are going to be solving oblique triangles Any triangle that is not a right triangle

In general: C a b A B c

To solve an oblique triangle, we must know 3 pieces of information: 1 Side of the triangle Any 2 other components Either 2 sides, an angle and a side, and 2 angles

C AAS ASA SSA SSS SAS Law of Sines a b A B c

Law of Sines If ABC is a triangle with sides a, b, and c, then:

ASA or AAS C A = 49º 27.4 102.3º a = 43.06 28.7º B A c =

ASA or AAS C A = 49º 27.4 102.3º a = 43.06 28.7º B A c = 55.75

Solve the following Triangle: A = 123º, B = 41º, and a = 10 C = 16º C 10 b 123º 41º c

C C = 16º 10 b = 7.8 b 123º 41º c

C C = 16º 10 b = 7.8 b c = 3.3 123º 41º c

Solve the following Triangle: A = 60º, a = 9, and c = 10 How is this problem different? C What can we solve for? 9 b 60º B 10

C C = 74.2º 9 b 60º B 10

C C = 74.2º 9 b B = 45.8º c = 7.5 60º B 10

What we covered: Solving right triangles using the Law of Sines when given: Two angles and a side (ASA or AAS) One side and two angles (SSA) Tomorrow we will continue with SSA

SSA The Ambiguous Case

Yesterday Yesterday we used the Law of Sines to solve problems that had two angles as part of the given information. When we are given SSA, there are 3 possible situations. No such triangle exists One triangle exists Two triangles exist

Consider if you are given a, b, and A Can we solve for h? a b h h = b Sin A A If a < h, no such triangle exists

Consider if you are given a, b, and A h A If a = h, one triangle exists

Consider if you are given a, b, and A h A If a > h, one triangle exists

Consider if you are given a, b, and A If a ≤ b, no such triangle exists

Consider if you are given a, b, and A If a > b, one such triangle exists

Hint, hint, hint… Assume that there are two triangles unless you are proven otherwise.

Two Solutions 31 12 20.5º Solve the following triangle. a = 12, b = 31, A = 20.5º 31 12 20.5º

2 Solutions B = 64.8º C = 94.7º c = 34.15 B’ = 180 – B = 115.2º First Triangle Second Triangle B = 64.8º C = 94.7º c = 34.15 B’ = 180 – B = 115.2º C’ = 44.3º C’ = 23.93

Problems with SSA Solve the first triangle (if possible) Subtract the first angle you found from 180 Find the next angle knowing the sum of all three angles equals 180 Find the missing side using the angle you found in step 3.

A = 60º; a = 9, c = 10 C = 74.2º B = 48.8º b = 7.5 C’ = 105.8º First Triangle Second Triangle C = 74.2º B = 48.8º b = 7.5 C’ = 105.8º B’ = 14.2º b ’ = 2.6

One Solution Solve the following triangle. What happens when you try to solve for the second triangle? a = 22; b = 12; A = 42º

a = 22; b = 12; A = 42º B = 21.4º C = 116.6º c = 29.4 B’ = 158.6º First Triangle Second Triangle B = 21.4º C = 116.6º c = 29.4 B’ = 158.6º C’ = -20.6º

No Solution Error → No such triangle Solve the following triangle. a = 15; b = 25; A = 85º Error → No such triangle

Law of Sines Section 6.1

Warm Up Solve the following triangles (if possible) A = 58º; a = 20; c = 10 B = 78º; b = 207; c = 210 A = 62º; a = 10; b = 12

A = 58º; a = 20; c = 10 C = 25º B = 97º b = 23.4

B = 78º; b = 207; c = 210 C = 82.9º A = 19.1º b = 69.2

B = 78º; b = 207, c = 210 C = 82.9º A = 19.1º a = 69.2 C’ = 97.1º First Triangle Second Triangle C = 82.9º A = 19.1º a = 69.2 C’ = 97.1º A’ = 4.9º a ’ = 18.1

A = 62º; a = 10; b = 12 No such triangle

Area of an oblique triangle

What is the area of a triangle? b b = c h h = b Sin A A

Area of an Oblique Triangle The area of any triangle is one-half the product of the lengths of two sides times the sine of their included angle. i.e.: Area = ½ bc Sin A = ½ ac Sin b = ½ ab Sin B

Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102º. Area = ½ (Side) (Side) Sine (angle) Area = ½ (90) (52) Sine (102º) Area = 1,189 m²

Find the area of the triangle: A = 35º, a = 14.7, b = 14.7

The course for a boat race starts at a point A and proceeds in the direction S 52º W to point B, then in the direction of S 40º E to point C, and finally back to A. Point C lies 8 miles directly south of point A. How far was the race? A 52 B 40 C

An airplane left an airport and flew east for 169 miles An airplane left an airport and flew east for 169 miles. Then it turned northward to N 32º E. When it was 264 miles from the airport, there was an engine problem and it turned to take the shortest route back to the airport. Find θ, the angle through which the airplane turned.

A farmer has a triangular field with sides that are 450 feet, 900 feet, and an included angle of 30º. He wants to apply fall fertilizer to the field. If it takes one 40 pound bag of fertilizer to cover 6,000 square feet, how many bags does he need to cover the entire field?