Section 6.5 The Law of Sines

Transcription

1 Setion 6.5 The Lw of Sines To solve tringle, we need to know ertin informtion out its sides nd ngles. A tringle is determined y three of its six prts (ngles nd sides) s long s t lest one of these three prts is side. So, the possiilities, illustrted in the Figures ove, re s follows. Cse One side nd two ngles (ASA or SAA) Cse Two sides nd the ngle opposite one of those sides (SSA) Cse 3 Two sides nd the inluded ngle (SAS) Cse 4 Three sides (SSS) Cses nd re solved using the Lw of Sines; Cses 3 nd 4 require the Lw of Cosines. The Lw of Sines The Lw of Sines sys tht in ny tringle the lengths of the sides re proportionl to the sines of the orresponding opposite ngles. Proof: To see why the Lw of Sines is true, refer to the Figure on the right. By the formul in Setion 6.3 the re of tringle ABC is sin C. By the sme formul the re of this tringle is lso nd. Thus, sin C Multiplying y /() gives the Lw of Sines. EXAMPLE: A stellite oriting the erth psses diretly overhed t oservtion sttions in Phoenix nd Los Angeles, 340 mi prt. At n instnt when the stellite is etween these two sttions, its ngle of elevtion is simultneously oserved to e 60 t Phoenix nd 75 t Los Angeles. How fr is the stellite from Los Angeles?

2 EXAMPLE: A stellite oriting the erth psses diretly overhed t oservtion sttions in Phoenix nd Los Angeles, 340 mi prt. At n instnt when the stellite is etween these two sttions, its ngle of elevtion is simultneously oserved to e 60 t Phoenix nd 75 t Los Angeles. How fr is the stellite from Los Angeles? Solution: Whenever two ngles in tringle re known, the third ngle n e determined immeditely euse the sum of the ngles of tringle is 80. In this se, C 80 ( ) 45 (see the Figure on the right), so we hve sin 60 sin C sin sin sin 45 3 { } The distne of the stellite from Los Angeles is pproximtely 46.4 mi. EXAMPLE: Solve the tringle in the Figure elow. Solution: Sine side is known, to find side we use the reltion sin C sin C 80.4 sin 0 sin To find, we first note tht B 80 (0 + 5 ) 35. Therefore sin C sin C 80.4 sin sin 5

3 The Amiguous Cse In the two previous Exmples unique tringle ws determined y the informtion given. This is lwys true of Cse (ASA or SAA). But in Cse (SSA) there my e two tringles, one tringle, or no tringle with the given properties. For this reson, Cse is sometimes lled the miguous se. To see why this is so, we show in the Figures elow the possiilities when ngle A nd sides nd re given. In prt () no solution is possile, sine side is too short to omplete the tringle. In prt () the solution is right tringle. In prt () two solutions re possile, nd in prt (d) there is unique tringle with the given properties. EXAMPLE: Solve tringle ABC, where A 45, 7, nd 7. Solution: We first find B. 7 ( ) ( ) 7 sin 45 So,, therefore B is either 30 or 50. Sine A 45, we nnot hve B 50, euse > 80. Hene B 30 nd the remining ngle is C 80 ( ) 05 Now we n find side. sin C sin C 7 sin 05 sin 30 7 sin 05 / 4 sin EXAMPLE: Solve tringle ABC, where A 43., 86., nd

4 EXAMPLE: Solve tringle ABC, where A 43., 86., nd Solution: From the given informtion we sketh the tringle shown in the Figure elow. Note tht side my e drwn in two possile positions to omplete the tringle. From the Lw of Sines 48.6 sin There re two possile ngles B etween 0 nd 80 suh tht Using lultor, we find tht one of these ngles is sin (0.95) The other is pproximtely We denote these two ngles y B nd B so tht B 65.8 nd B 4. Thus two tringles stisfy the given onditions: tringle A B C nd tringle A B C. Solve tringle A B C : C 80 ( ) 7. Thus Solve tringle A B C : sin C 86. sin 7. sin Thus C 80 ( ).7 sin C 86. sin.7 sin Tringles A B C nd A B C re shown in the Figures elow. EXAMPLE: Solve tringle ABC, where A 4, 70, nd. 4

5 EXAMPLE: Solve tringle ABC, where A 4, 70, nd. Solution: We hve sin Sine the sine of n ngle is never greter thn, we onlude tht no tringle stisfies the onditions given in this prolem. 5