Mth-1060 Lesson 4-5 The Lw of osines
Solve using Lw of Sines. 1 17 11 5 15 13 SS SSS Every pir of loops will hve unknowns. Every pir of loops will hve unknowns. We need nother eqution.
h Drop nd ltitude from Vertex to the opposite side. sin h sin h sin Sustitute sin for h in the figure.
sin sin x os x os os x os
sin os os Using Pythgoren Theorem on the right-side tringle: ( sin ) ( os sin sin os (sin os os 1 ) os ) os os Lw of osines: Uses 3 sides nd 1 ngle.
Lw of osines os os os Then wy. in the usul e ny tringle with sides nd ngles leled Let os os os 1 1 1 Pythgoren Theorem
There is pttern for Lw of osines 1. Lel the tringle.. Loop the 1 side nd its opposite ngle nd the other sides. If 3 of the 4 vlues re known, use Lw of osines.
Lw of osines os If 90 (right tringle) Notie in the pttern tht ngle is opposite ends of the eqution from side length. os 90? 0 os Is tht fmilir? eomes
Lw of osines: Ptterns in the formul os os os
There re some Goth s with the Lw of osines (when there is n otuse ngle). Lw of osines will lwys e the 1 st step for SSS nd SS. When solving for the other sides nd ngles, you n hoose etween Lw of osines (more diffiult lultion) or Lw of Sines (esier lultion). 0.7071 0. 7071 sin(135)? sin(45)? When using the Lw of Sines, the lultor will give the smllest ngle even if the ngle is tully otuse. If you use Lw of osines for follow-on steps, there is no miguity. You ll get the orret ngle. 0. os(135)? 7071 0. os(45)? 7071 If you hoose to use the Lw of Sines, find the smllest ngle 1 st.
For SSS, you n find ny ngle 1 st. 7 5 4 4 7 5 os 1 (0.886) ( os Reple letters with prentheses. ) ( ) ( ) ( )( )os( Sustitute numers into the prentheses. (4) 34 ( 7) (5) os ) (7)(5) os( ) 16 49 5 70os 16 74 70os n we sutrt 70 from 74? 16 74 os 70 0.886 os
You n use Lw of osines Next 7 4 34 S 5 sin 5 7 4 34 5 sin 34 4 sin 1 ( (5) ) ( ( ) 7) ( ) (4) (5 7 4 ) os ( *7*4) os 1 (0.7143) Or Lw of Sines 5sin 34 4 44.4 SS 5 ( os )( )os( ) (7)(4) os( ) sin 7 0.7143 44.4 7 4 os sin 34 4 7sin 34 sin 1 4 101.6 34 67. 9 Why re they different? 78.1
For SS, you must find the missing side 1 st. 0 5 11 5 0º 11 (1) If it is not lredy given, drw nd lel tringle. () Wht pttern? Lw of Sines? Or Lw of osines? (3) Use Lw of osines 5 11 (5)(11) os 0 5 11110os 0 4.6 6.5
5 0º 11 6.5 15.3º sin (5) Tringle Sum Theorem (4) Whih ngle do you find next? 5sin 0 6.5 m m m 180 m 180 0 15.3 m 144.7 Lw of Sines is esier. Find the smllest ngle first!!!!) sin sin sin 5 5sin 0 sin 1 6.5 sin 0 6.5 15.3 Wht ngle would Lw of Sines give you if you found ngle in step (4) ove? 180 144.7 35.3 These two ngles hve the sme referene ngle. Whih is the orret ngle?
For the SSS se: find the lrgest ngle first! (This will get the otuse ngle out in the open if it exists in the tringle, so susequent use of Lw of Sines won t hve the goth. ) 1 If you don t find the otuse zº ngle 1 st you hve to rememer 0 to find the smllest ngle next yº when you use the Lw of Sines. 7 xº 144 119 x os 1 1080 144 119 x os 1 1080 x 4. (1) (0) (7) os (0)(7)os(x) 144 400 79 1080os(x ) 144 119 1080os(x )
We didn t find the otuse ngle 1 st using Lw of osines so, if you use the Lw of Sines to find the nd ngle, find the SMLLEST NGLE next. 0 4.º zº 7 1 yº sin y z sin y sin sin 0 0sin 4. 1 0sin 4. sin 1 1 180 4. y 43.1 y sin 4. 1 z 43.1 11.7 Wht ngle would Lw of Sines give you if you found ngle z ove? 180 11.7 67.3 These two ngles hve the sme referene ngle. Whih is the orret ngle?
ltitude of tringle: The perpendiulr distne from ny vertex to its opposite side. ltitude of tringle: mens the sme thing s the height of tringle. Height ltitude How mny heights (ltitudes) does tringle hve? se of tringle: ny side of tringle. Height ltitude 1 se side * se* height three How mny different wys re there to lulte the re of tringle? three
1 * se* height re formul: requires the use of mthing heights nd sides. Using segment s the se, requires the use of segment E s the height.
There re two right tringles tht n e used to solve for h. sin sin( 48) opp hyp h 8 h 8*Sin(48º) re 0.5* se* height 7.6 81º h 8 h 5.9 re ½(10)(5.9) 51º 10 48º re 9.5 squre units
For height h we ould use the other tringle. sin opp hyp re 0.5* se* height sin(51) h 7.6 h 7.6*Sin(51º) h 5.9 7.6 81º h 8 re ½(10)(5.9) 51º 10 48º re 9.5 squre units Either of the two tringles gives us the sme height.
6.1 4. 44 6. 3 Find the re of the tringle using ltitude E s its height. 1 * se* height 1 **E 1 *(6.3)*(4.08) 1.89 units E 6.1 sin 41.8 E 6.1sin 41.8 E 4.08 E 4.44 sin 66.63 E 4.44sin 66.63 E 4.08
3 re 5 1 *se*height Wht is the se? Wht is the height? 4 For right tringles one of the legs is the se, the other leg is the height (or vie-vers). There is no point in finding the height from ngle to side.
15, 7, 5, Find the re of the Tringle (Two sides nd n ngle, ll 3 letters re different SS) 5 0.5* se* height 0.5*(lt.)* 7 15º lt 5 sin 15 lt 5 sin 15 lt 1.9 0.5 1.9 7 4.5 units
Wht is the re of given tht 5, 3, nd 14. (1) Find n ngle! SSS Lw of osines (find ny ngle). os 14 3 5 (3)(5)os 3 5 os 1 (14 3 5 ( *3*5) ) 14 33.6
Wht is the re of given tht 5, 3, nd 14. () Using ngle, find the pproprite height.. sin 33.6 h 5 3 33.6º 5 h 5sin 33.6 h 13.8 h re 0.5* se* height 14 re 0.5(3)(13.8) re 159 units
Let s look t it gin. SIDE NGLE SIDE 3 33.6º 5 re 0.5* se* height re 0.5(3)(5sin33.6) re 0.5***sin Notie: in the formul you see ll three letters,, nd h13.8 14
Deriving Heron s Formul SIDE NGLE SIDE opp sin hyp h sin h sin h Sustitute into re formul 1 re *se*height re ½***sin Notie: in the formul you see ll three letters,, nd re ½***sin re ½***sin
y omining the Lw of Sines nd the SS Tringle re formul (followed y lot of steps), we n derive formul for the re of SSS tringle. s(s )(s )(s ) s Where s is the semi-perimeter (1/ perimeter) of the tringle. Find the re of the tringle. s 14 3 5 31 31(31 14)(31 3)(31 5) 3 5 159 units 14