6.1: The Law of Sines

Law of Sines In any triangle ABC (in standard notation): a__ = b__ = c__ Sin A Sin B Sin C * Used with AAS and SSA problems

Example If B = 20˚, C = 31˚, and b = 210, find the other angle measure and side lengths. C AB 31˚ 20˚ 210 a c <A = 180 – ( ) = 129˚ a__ = 210_ Sin 129 Sin 20 a = c__ = 210_ Sin 31 Sin 20 c =

Practice If B = 97.5˚, C = 42.5˚, and b= 7, find the other angle measure and side lengths. m<A = 40 a = 4.54 c = 4.77

Ambiguous Case (SSA) May be 1, 2, or no triangles that satisfy the case. Possible because SSA is not a congruence theorem. Open book to page 430. A < 90 there are 4 possibilities A> 90 there are 2 possibilities

Supplementary Angle Identity If 0˚ < θ < 90˚, then Sin θ = sin (180˚ - θ)

Example Given a possible triangle ABC with a = 6, b = 7, and m<A = 65˚, find angle B. A BC ___ = 7___ Sin 65 sin θ 6 sin θ = 7Sin 65 sin θ = 7Sin 65 6 sin θ = 1.06 Angle does not exist

Practice Given a possible triangle ABC with a = 12.4, c = 6.2, and A = 72˚, find angle C. m<C = 28.40˚ No Other Options because…… 180 – = = > 180 (sum of a triangle)

Example Solve triangle ABC when a = 7.5, b = 12, and A = 35˚. Lets find the angles first. 7.5 = 12__ Sin 35 sin B B = or B = C is : 180 – – 35 = or 180 – – 35 = 31.6

Example Solve triangle ABC when a = 7.5, b = 12, and A = 35˚. Now lets find side c. 7.5 = x__ Sin 35 sin c = or c = = x__ Sin 35 sin 31.6

Practice Solve triangle ABC when a = 30, b= 40, and A = 30˚. m<B = ˚ or ˚ m<C = ˚ or 11.81˚ c = 57 or

Area of a triangle – (SAS) The area of a triangle containing an angle C with adjacent sides of lengths a and b is: (1/2)(a)(b)(Sin C)

Example Find the area of the triangle shown. 8 cm 13 cm 130˚ (1/2)(8)(13)(Sin 130) =39.83 cm 2

Practice Find the area of the triangle shown ˚ (1/2)(7)(10)(Sin 68) =32.45

Heron’s Formula – (SSS) The area of a triangle with sides a, b, and c is: √(s(s – a)(s – b)(s – c)) Where s = ½ (a + b + c)

Example Find the area of the triangle whose sides have lengths 7, 9, and 12. s = ½( ) =14 √(14(14 – 7)(14 – 9)(14 – 12)) =31.31

Practice Find the area of the triangle whose sides have lengths 4, 12, and 14. s = ½( ) =15 √(15(15 – 4)(15 – 12)(15 – 14)) =22.25

Assignment Pg – 14 even, 18, 26 – 34 even, 40 – 44 even