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6.2 Law of Cosines

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6.2 Law of Cosines 6.2 Law of Cosines The Law of Sines can’t be used directly to solve triangles if we know two sides and the angle between them or if we know all three sides. In this two cases, the Law of Cosines applies. Law of Cosines: In any triangle ABC , we have a2 b 2 c 2 2 bc cos A b2 a 2 c 2 2 ac cos B c2 a 2 b 2 2 ab cos C Proof: To prove the Law of Cosines, place triangle so that A is at the origin, as shown in the Figure below. The coordinates of the vertices BC and are (c ,0) and (b cos A , b sin A ) , respectively. Using the Distance Formula, we have a2( c b cos) A 2 (0 b sin) A 2 =c2 2bc cos A b 2 cos 2 A b 2 sin 2 A =c2 2bc cos A b 2 (cos 2 A sin 2 A ) =c22 2bc cos A b =b22 c 2 bc cos A Example: A tunnel is to be built through a mountain. To estimate the length of the tunnel, a surveyor makes the measurements shown in the Figure below. Use the surveyor’s data to approximate the length of the tunnel. Solution: c2 a 2 b 2 2 ab cos C 21222 388 2 212 388cos82.4 173730.23 c 173730.23 416.8 Thus, the tunnel will be approximately 417 ft long. Example: The sides of a triangle are a5, b 8, and c 12. Find the angles of the triangle. Solution: a2 b 2 c 2 2 bc cos A b2 c 2 a 28 2 12 2 5 2 cosA 0.953125 2bc 2 8 12 A cos1 (0.953125) 18 Similarly, you can get BC 29 , 133 . Remark: You can only compute two angles using the Law of Cosines, then the third angle, you can get by the relationship of the sum of three angles is 180 . Example: Solve triangle ABC , where A 46.5 , b 10.5 and c 18 . Solution: We can use the Law of Cosines to find a first. a2 b 2 c 2 2 bc cos A 10.5 2 18 2 2 10.5 18cos46.5 174 a 174 13.2 a2 c 2 b 213.2 2 18 2 10.5 2 cosB 0.816477 2ac 2 13.2 18 B cos1 (0.816477) 35.3 a2 b 2 c 213.2 2 10.5 2 18 2 cosC 0.142532 2ab 2 13.2 10.5 C cos1(0.142532 ) 98.2 The area of triangle. Heron’s Formula. The area of triangle ABC is given by area s( s a )( s b )(s c) 1 where, s( a b c ), is the semiperimeter of the triangle, that is, s is half the perimeter. 2 1 Proof: We have the formula area absin C in last section. Thus, 2 1 area2 a 2 b 2sin 2 C 4 1 =a2 b 2 (1 cos 2 C ) 4 1 =a22 b (1 cos C )(1 cos C ) 4 By the Law of Cosines, we have a2 b 2 c 2 cosC 2ab a2 b 2 c 2 1 cosC 1 2ab 2ab a2 b 2 c 2 2ab ()a b22 c 2ab (a b c )( a b c ) 2ab Similarly, (c a b )( c a b ) 1-cosC= 2ab 1 area2 a 2 b 2sin 2 C 4 1 =a2 b 2 (1 cos 2 C ) 4 1 (cabcab )( ) ( abcabc )( ) = ab22 4 2ab 2 ab abcabcacbbca = 2 2 2 2 =s ( s c )( s b )( s a ) Thus, we have area s( s c )( s b)( s a ) Example: A businessman wishes to buy a triangular lot in a busy downtown location (see the Figure below). The lot frontages on the three adjacent streets are 125, 280, and 315 ft. Find the area of the lot. Solution: 125 280 315 s 360 2 By Heron's Formula the area is area=s ( s a )( s b )( s c )= 360(360 125)(360 280)(360 315) 17452 Thus, the area is approximately 17,45 ft 2 .
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