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> Zdatak Iz Analiticke Geometrije, zdatak iz analiticke geometrije

vriskica
post Aug 23 2008, 02:13 PM
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ne izadje mi dobro pa hajd vi probajete
Zadan je trougao ABC i tacka M kao poloviste stranice BC,gdje je IABI=2 , IACI=3 a ugao <BAC=pi / 3.Izraziti vektore AM i BC pomocu AC i AB.Izracunati ugao izmedju BC i AM i povrsinu trougla ABM.

rijesenje
Da resis trougao koristi kosinusnu teoremu (imas dve stranice i ugao izmedju njih).
Onda lako mozes da nadjes ostale uglove u trouglu a povrsina ABM ti je polovina povrsine ABC jer je M simetrala (srediste) stranice.
Da nadjes povrsinu koristi Heronov obrazac.

Vektori:
AM=AB+BM
AM=AC+CM
BM=-CM
Ako saberemo prvu i drugu j-nu:
2AM=AB+AC

I
BC=BA+AC=-AB+AC
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VIP <3
post Aug 23 2008, 02:20 PM
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Eeeeeeeeeee da sam znala da je analiticka geometrija ovako laka mozda bih je bolje prihvaitla u 3. razredu XD.gif Hm, ovo je stvarno analiticka? Ja sam mislila da je analiticka ono Bozino nerazumno sa kapicama XD.gif

Edit: Ne razumem koji deo mi treba da resimo, tj sta ti nije jasno? unsure.gif

This post has been edited by VIP (ex Miss): Aug 23 2008, 02:23 PM


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vriskica
post Aug 23 2008, 02:38 PM
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ma had uradite tako kako sam ja ..pa da vidim kako ce kod vas ispast..i koliki ce brojevi izac..molim vas treba mi hitno ..
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pyost
post Aug 23 2008, 02:49 PM
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Dobro je ovo sto si napisala

2AM = AB + AC
BC = BA + AC = AC - AB


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vriskica
post Aug 23 2008, 02:52 PM
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ako znas uradi cijeli zadatak..pa da vidim tvoje rijesenje..unaprijed hvala
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pyost
post Aug 23 2008, 02:59 PM
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Ja ne mogu, na moru sam laugh.gif Ali naci ce se neko wink.gif


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vriskica
post Aug 23 2008, 03:00 PM
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ok ...valjda ce se naci neko..
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Wolfbrother
post Aug 23 2008, 04:23 PM
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a^2=b^2+c^2-2\cos\alpha bc \rightarrow a=\sqrt 7
Iz Stjuartove teoreme http://en.wikipedia.org/wiki/Stewart_theorem za x=y imamo b^2a/2+c^2a/2=a^3/4+p^2a \rightarrow p=\sqrt{19} /2
Sada primenimo kosinusnu na  \triangle AMB:  c^2=p^2+a^2/4-2pa\cos\varphi  \rightarrow \cos\varphi=5/\sqrt{133}   \sin\varphi=\sqrt{ \frac{108} {133} } P\triangle AMB=sin\varphi pa/4=\sqrt{3}*3/4
Valjda nisam pogresio nista, mnogo su ruzni brojevi. Al u svakom slucaju svakako je bolje naci ovu povrsinu preko sinusa nego heronovim obrascem.

This post has been edited by Wolfbrother: Aug 23 2008, 05:29 PM


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vriskica
post Aug 23 2008, 04:46 PM
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HVALA PUNO..
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VIP <3
post Aug 23 2008, 04:49 PM
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Ae Wolfbrother objasni mi sto si je gnjavio uvodjenjem za nju vrvtno novog pojma ( Stjuartovom teoremom)? Ona je samo htela brojeve da joj neko proveri, cak je i napisala da ce koristiti kosinusnu i Heronov obrazac dry.gif
Mene je mrzelo da racunam, izvini vriskice, ne uzimam papir i olovku u ruke pre 1. septembra wink.gif


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Wolfbrother
post Aug 23 2008, 04:51 PM
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Tako mi bilo najlakse da skrljam, uostalom stjuartova se dokazuje preko dve kosinusne pa i nije jelte nekakvo cudoviste. smile.gif
Ali ako nisam omasio u racunu ono bi se razbio od racunanja da resis heronovim, malo je nelogicno to tako raditi.

This post has been edited by Wolfbrother: Aug 23 2008, 04:52 PM


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VIP <3
post Aug 23 2008, 04:58 PM
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Ali ona tako hoceeeeeeeeeeeeeeeeeeeeeeeeeeee dry.gif Mislim, mozda (ne kazem da je tako, ali MOZDA) ona recimo sprema..... hm popravni iz necega... vrvtno ne, ali nije ni bitno.... i onda je malo cudno da se pojavi i profesoru kaze: "Setimo se Stjuartove teoreme" XD.gif Mislim, ne znam jesi me razumeo... prosto, devojka (pp da je zensko XD.gif ) je to (valjda) s nekim svojim razlogom htela tako XD.gif A mozda i nije XD.gif Ko ce ga znati XD.gif


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vriskica
post Aug 23 2008, 05:00 PM
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HVALA TI PUNO ali ako neko zna uradit preko herona..i nije mi jasan ovaj ugao izmedju BC I AM..
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Wolfbrother
post Aug 23 2008, 05:02 PM
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Napisao sam ga kao \varphi, a preko herona ti je prakticno isto...mislim imas trougao i sve njegove stranice...

This post has been edited by Wolfbrother: Aug 23 2008, 05:02 PM


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Wolfbrother
post Aug 23 2008, 05:03 PM
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QUOTE(VIP (ex Miss) @ Aug 23 2008, 05:58 PM)
Ali ona tako hoceeeeeeeeeeeeeeeeeeeeeeeeeeee dry.gif Mislim, mozda (ne kazem da je tako, ali MOZDA) ona recimo sprema..... hm popravni iz necega... vrvtno ne, ali nije ni bitno.... i onda je malo cudno da se pojavi i profesoru kaze: "Setimo se Stjuartove teoreme" XD.gif Mislim, ne znam jesi me razumeo... prosto, devojka (pp da je zensko XD.gif ) je to (valjda) s nekim svojim razlogom htela tako XD.gif A mozda i nije XD.gif Ko ce ga znati XD.gif
*


Ne znam sta da ti kazem nikad nisam spremao popravni. XD.gif
 s=(x+y+z)/2  P=\sqrt{s(s-x)(s-y)(s-z)}

x=sqrt{19}/2 y=sqrt{7}/2 z=2

Mada trebalo bi opet proveriti brojeve...

This post has been edited by Wolfbrother: Aug 23 2008, 05:12 PM


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VIP <3
post Aug 23 2008, 05:10 PM
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Nisi nikad pomagao drugu koji je okinuo, ili mu makar resavao neke zadatke kad se sprema u junu da popravlja keca? Ono kad dobijes gomilu banalnih zadataka, od cega ce on imati na testu recimo 5, ali moras da ih uradis na toliko banalan nacin (jer je to ono sto on kao zna XD.gif ) da ti se zivot smuci od ispisivanja necega sto ti i sam vidis i od nemogucnosti da preskocis po koji red? XD.gif Awwwwwwww pa ti i ne znas kakvu zabavu si propustio XD.gif


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Wolfbrother
post Aug 23 2008, 05:30 PM
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Damn, nesto sam pogresio u racunu kod kosinusa i valjda sad ispravio, nemam vremena da proverim sad jer idem kod zubara, ali u svakom slucaju moze ovako da se uradi.
Herovim obrascem bi moglo i P\triangle ABC=sqrt{(5+\sqrt7)(1+\sqrt7)(\sqrt7-1)(5-\sqrt7)}/4=\sqrt{108}/4

This post has been edited by Wolfbrother: Aug 23 2008, 05:37 PM


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