13. Law of Sines

Introduction In this section, we will solve (find all the sides and angles of) oblique triangles – triangles that have no right angles. As standard notation, the angles of a triangle are labeled A, B, and C, and their opposite sides are labeled a, b, and c. To solve an oblique triangle, we need to know the measure of at least one side and any two other measures of the triangle—either two sides, two angles, or one angle and one side.

4 cases Two angles and any side are known (AAS or ASA) Two sides and an angle opposite one of them are known (SSA) Three sides are known (SSS) Two sides and their included angle are known (SAS) The first two cases can be solved using the Law of Sines, the last two cases require the Law of Cosines.

A S A ASA S A A SAA CASE 1: ASA or SAA Law of sines

S A S CASE 2: SSA Law of sines

S A S CASE 3: SAS Law of cosines

S S S CASE 4: SSS Law of cosines

Deriving the Law of Sines Triangle ABC does not have a right angle so we drop a perpendicular from Angle B We call this h and now we have a right angle so we can use SOHCAHTOA B c a h A C b sin A=h/c, h = csinA sin C=h/a, h = asinC csinA=asinC

Law of Sines A b c C B a

Remember ……

In a triangleC = 102, B = 29, and b = 28 feet. Solve the triangle Case 1 - AAS In a triangleC = 102, B = 29, and b = 28 feet. Solve the triangle

Example AAS - Solution The third angle of the triangle is A = 180 – B – C = 180 – 29 – 102 = 49. By the Law of Sines, you have .

Example AAS – Solution cont’d Using b = 28 produces and

Case 1 - ASA A B C c a b C=70o b=44.1 a=32.7

Area of an Oblique Triangle (Given SAS)

Area of a Triangle Area = ½ bh sin A = h/c h = c sin A Area = ½ b c sin A

Area of a Triangle Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A) c A h Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

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.

Try It Out Determine the area of these triangles 42.8° 127° 76.3° 17.9 24 12 76.3°