7. SOLVING OBLIQUE TRIANGLES: THE LAW OF SINES

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1 7 SOLVING OLIQUE TRINGLES: THE LW OF SINES n ique tringe is ne withut n nge f mesure 90 When either tw nges nd side re knwn (S) in the tringe r tw sides nd the nge ppsite ne f them (SS) is given, then the w f sines my e ppied t sve the tringe Therem 71: The Lw f Sines In the gener tringe sin sin β sin γ T prve the w f sines fr the ique tringe shwn in Figure 71, we wi first shw tht sin sin β γ β Figure 71 Fr this, we drp perpendiur D frm vertex t the stright ine ntining the verties nd euse th nges nd β re ute in the figure, D ies etween nd In the right tringes D nd D, we hve sin D D nd sin β s sin D nd sin β D nsequenty, sin sin β Dividing th sides f the st equtin y, we tin sin sin β 46

2 E γ h D β Figure 72 Nw t verify tht sin sin γ, we serve tht nge γ is tuse Let E e the perpendiur frm vertex t the ine ntining side Let Then h h sin γ sin ( 180 γ ) sin γ nd sin s sin γ sin sin sin γ Thus, The prf f the w f sines in the se f n ute tringe is simir E h Ntie tht the w f sines (Therem 71) n e written in the terntive frm: sin sin β sin γ Sving Tringe (S) If tw nges f the tringe re given, then the third nge n e fund y using the retinship: + β + γ 180 ; 47

3 hene, the three denmintrs sin, sin β, nd sin γ n e fund using utr Nw, if ny ne f the sides,, r is s given, then the equtins sin sin β sin γ n e sved fr the remining tw sides The fwing exmpe indites the predure fr sving tringe when tw nges nd ne side re given (r n e determined frm the infrmtin prvided) Uness therwise indited, we sh rund ff nges t the nerest hudredth f degree, nd side engths t fur signifint digits Exmpe In, suppse tht 41, β 77, nd 74 Sve fr γ,, nd β Figure γ Sutin γ y the w f sines, hene, sine 74, nd sin 77 sin sin 77 74sin 62 ; sin 77 sin Nte tht in this se there is wys unique sutin y the nge-side-nge riteri fr ngruent tringes Sving Tringe (SS) euse there re sever pssiiities, the situtin in whih yu re given the engths f tw sides f tringe nd the nge ppsite ne f them is ed the miguus se Fr instne, suppse yu re given side, side, nd nge in Yu might try t nstrut frm this infrmtin y drwing ine segment f ength nd ry tht strts t nd mkes n nge with (Figure 74) T find the 48

4 remining vertex, yu ud use mpss t drw n r f ire f rdius with enter If the r intersets the ry t pint, then is the desired tringe γ Figure 74 β I s figure 7 iustrtes, there re tuy fur pssiiities if yu try t nstrut y the ve methd in se is ute: ( i ) The ire des nt interset the ry t nd there is n tringe (Figure 7) ( ii ) The ire intersets the ry in exty ne pint nd there is just ne right tringe (Figure 7) ( iii ) The ire intersets the ry in tw pints 1 nd 2 nd there re tw tringes 1 nd 2 (Figure 7) ( iv ) The ire intersets the ry in exty ne pint nd there is just ne ute tringe (Figure 7d) Figure 7 ( ) < sin n tringe pssie ( ) sin ne right tringe 49

5 1 2 ( ) sin < < tw tringes ( d ) > ne ute tringe II In se is tuse, then there re ny tw pssiiities s shwn in Figure 76 Figure 76 ( ) > ne tringe ( ) < n tringe pssie In the miguus se, yu n wys use utr t sve the tringe Just use the w f sines, sin sin β sin t evute sin β : sin β, 0 < β < 180 0

6 sin Re tht the sine f n nge is never greter thn 1; hene, if > 1, then this trignmetri equtin hs n sutin, nd n tringe stisfies the given nditins If sin sin 1, then the equtin hs ny ne sutin, 90 If < 1, then the trignmetri equtin hs tw sutins Nmey, sin 1 rsin nd One yu hve determined ( r 1 nd 2 ), yu knw tw nges nd tw sides f the tringe (r tringes), nd yu n sve the tringe y using the methds previusy expined Even if there re tw sutins 1 nd 2 f the trignmetri equtin fr, it is pssie tht ny ne f these sutins rrespnds t n tu tringe stisfying the given nditins (see Figure 7d nd Figure 76) Exmpe Sve the tringe in eh se 1 30, 8, Sine is ute nd > there is ny ne ute tringe nstrutie sin sin sin rsin Then sin sin 30 Finy, sin sin ,, 8 Here is ute nd sin 2 8 < < there re tw tringes nstrutie sin 30 sin sin

7 Thus, β 1 rsin 4 sin Then γ 1 sin 9687 T find the mesure f β 2 serve tht nd sin 9687 sin sin β 2 sin ( π β1 ) sinπ s β 1 s π sin β 1 sin β 1 Thus, β β nd γ sin30 sin 2313 sin 2313 Then sin30 Setin 7 Prems In prems 1 t 10 use the w f sines t sve eh tringe Rund ff nges t the nerest hundredth f degree nd side engths t fur signifint digits 1 32, β 38, γ , 30, , 70, , 9, γ 60 31, 33, 60 6 β 14, , 8, 8 30,, 8 9 β 60, 11, β 60, 12, 11 2

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