The Law of SINES

Working with Non-right Triangles We wish to solve triangles which are not right triangles B A C a c b h Drawing a different altitude will allow to prove similarly these both =

The Law of SINES For any triangle (right, acute or obtuse), you may use the following formula to solve for missing sides or angles:

AAS - 2 angles and 1 adjacent side Use Law of SINES when ... you have 3 dimensions of a triangle and you need to find the other 3 dimensions - they cannot be just ANY 3 dimensions though, or you won’t have enough info to solve the Law of Sines equation. Use the Law of Sines if you are given: AAS - 2 angles and 1 adjacent side ASA - 2 angles and their included side SSA (this is an ambiguous case)

Example 1 You are given a triangle, ABC, with angle A = 70°, angle B = 80° and side a = 12 cm. Find the measures of angle C and sides b and c. * In this section, angles are named with capital letters and the side opposite an angle is named with the same lower case letter .*

Example 1 (con’t) A C B 70° 80° a = 12 c b The angles in a ∆ total 180°, so angle C = 30°. Set up the Law of Sines to find side b:

Example 1 (con’t) A C B 70° 80° a = 12 c b = 12.6 30° Set up the Law of Sines to find side c:

Example 1 (solution) A C B 70° 80° a = 12 c = 6.4 b = 12.6 30° Angle C = 30° Side b = 12.6 cm Side c = 6.4 cm Note: We used the given values of A and a in both calculations. Your answer is more accurate if you do not used rounded values in calculations.

Example 2 You are given a triangle, ABC, with angle C = 115°, angle B = 30° and side a = 30 cm. Find the measures of angle A and sides b and c.

Example 2 (con’t) To solve for the missing sides or angles, we must have an angle and opposite side to set up the first equation. We MUST find angle A first because the only side given is side a. The angles in a ∆ total 180°, so angle A = 35°. A C B 115° 30° a = 30 c b

Example 2 (con’t) A C B 115° a = 30 c b 30° a = 30 c b 35° Set up the Law of Sines to find side b:

Example 2 (con’t) A C B 115° a = 30 c b = 26.2 30° a = 30 c b = 26.2 35° Set up the Law of Sines to find side c:

Example 2 (solution) Angle A = 35° Side b = 26.2 cm Side c = 47.4 cm A 115° 30° a = 30 c = 47.4 b = 26.2 35° Angle A = 35° Side b = 26.2 cm Side c = 47.4 cm Note: Use the Law of Sines whenever you are given 2 angles and one side!

The Ambiguous Case (SSA) When given SSA (two sides and an angle that is NOT the included angle) , the situation is ambiguous. The dimensions may not form a triangle, or there may be 1 or 2 triangles with the given dimensions. We first go through a series of tests to determine how many (if any) solutions exist.

The Ambiguous Case (SSA) In the following examples, the given angle will always be angle A and the given sides will be sides a and b. If you are given a different set of variables, feel free to change them to simulate the steps provided here. ‘a’ - we don’t know what angle C is so we can’t draw side ‘a’ in the right position A B ? b C = ? c = ?

The Ambiguous Case (SSA) Situation I: Angle A is obtuse If angle A is obtuse there are TWO possibilities If a ≤ b, then a is too short to reach side c - a triangle with these dimensions is impossible. If a > b, then there is ONE triangle with these dimensions. A B ? a b C = ? c = ? A B ? a b C = ? c = ?

The Ambiguous Case (SSA) Situation I: Angle A is obtuse - EXAMPLE Given a triangle with angle A = 120°, side a = 22 cm and side b = 15 cm, find the other dimensions. Since a > b, these dimensions are possible. To find the missing dimensions, use the Law of Sines: A B a = 22 15 = b C c 120°

The Ambiguous Case (SSA) Situation I: Angle A is obtuse - EXAMPLE Angle C = 180° - 120° - 36.2° = 23.8° Use Law of Sines to find side c: A B a = 22 15 = b C c 120° 36.2° Solution: angle B = 36.2°, angle C = 23.8°, side c = 10.3 cm

The Ambiguous Case (SSA) Situation II: Angle A is acute If angle A is acute there are SEVERAL possibilities. Side ‘a’ may or may not be long enough to reach side ‘c’. We calculate the height of the altitude from angle C to side c to compare it with side a. A B ? b C = ? c = ? a

The Ambiguous Case (SSA) Situation II: Angle A is acute First, use SOH-CAH-TOA to find h: A B ? b C = ? c = ? a h Then, compare ‘h’ to sides a and b . . .

The Ambiguous Case (SSA) Situation II: Angle A is acute If a < h, then NO triangle exists with these dimensions. A B ? b C = ? c = ? a h

The Ambiguous Case (SSA) Situation II: Angle A is acute If h < a < b, then TWO triangles exist with these dimensions. A B b C c a h A B b C c a h If we open side ‘a’ to the outside of h, angle B is acute. If we open side ‘a’ to the inside of h, angle B is obtuse.

The Ambiguous Case (SSA) Situation II: Angle A is acute If h < b < a, then ONE triangle exists with these dimensions. Since side a is greater than side b, side a cannot open to the inside of h, it can only open to the outside, so there is only 1 triangle possible! A B b C c a h

The Ambiguous Case (SSA) Situation II: Angle A is acute If h = a, then ONE triangle exists with these dimensions. A B b C c a = h If a = h, then angle B must be a right angle and there is only one possible triangle with these dimensions.

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 Given a triangle with angle A = 40°, side a = 12 cm and side b = 15 cm, find the other dimensions. Find the height: A B ? 15 = b C = ? c = ? a = 12 h 40° Since a > h, but a< b, there are 2 solutions and we must find BOTH.

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 FIRST SOLUTION: Angle B is acute - this is the solution you get when you use the Law of Sines! A B 15 = b C c a = 12 h 40°

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse - use the first solution to find this solution. In the second set of possible dimensions, angle B is obtuse, because side ‘a’ is the same in both solutions, the acute solution for angle B & the obtuse solution for angle B are supplementary. Angle B = 180 - 53.5° = 126.5° a = 12 A B 15 = b C c 40° 1st ‘a’ 1st ‘B’

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse Angle B = 126.5° Angle C = 180°- 40°- 126.5° = 13.5° a = 12 A B 15 = b C c 40° 126.5°

The Ambiguous Case (SSA) Situation II: Angle A is acute - EX. 1 (Summary) Angle B = 53.5° Angle C = 86.5° Side c = 18.6 Angle B = 126.5° Angle C = 13.5° Side c = 4.4 a = 12 A B 15 = b C c = 4.4 40° 126.5° 13.5° A B 15 = b C c = 18.6 a = 12 40° 53.5° 86.5°

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 2 Given a triangle with angle A = 40°, side a = 12 cm and side b = 10 cm, find the other dimensions. A B ? 10 = b C = ? c = ? a = 12 h 40° Since a > b, and h is less than a, we know this triangle has just ONE possible solution - side ‘a’opens to the outside of h.

The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 2 Using the Law of Sines will give us the ONE possible solution: A B 10 = b C c a = 12 40°

The Ambiguous Case - Summary if angle A is acute find the height, h = b*sinA if angle A is obtuse if a < b  no solution if a > b  one solution if a < h  no solution if h < a < b  2 solutions one with angle B acute, one with angle B obtuse if a > b > h  1 solution If a = h  1 solution angle B is right (Ex I) (Ex II-1) (Ex II-2)

The Law of Sines AAS ASA SSA (the ambiguous case) Use the Law of Sines to find the missing dimensions of a triangle when given any combination of these dimensions.

Area of an Oblique Triangle

Area of an Oblique Triangle The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle. Referring to the triangles below, that each triangle has a height of h = b sin A. A is acute. A is obtuse.

Area of a Triangle - SAS B a h C Looking at this from all three sides: SAS – you know two sides: b, c and the angle between: A Remember area of a triangle is ½ base ● height Base = b Height = c ● sin A  Area = ½ bc(sinA) A B C c a b h Looking at this from all three sides: Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

Area of an Oblique Triangle

Example – Finding the Area of a Triangular Lot Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102. Solution: Consider a = 90 meters, b = 52 meters, and the included angle C = 102 Then, the area of the triangle is Area = ½ ab sin C = ½ (90)(52)(sin102)  2289 square meters.