Točno
13. siječnja 2016. 18:50 (8 godine, 6 mjeseci)
U paralelogramu
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
simetrala kuta
![\sphericalangle DAB](/media/m/f/d/8/fd81f94f045fb166610c69d8ae506717.png)
raspolavlja dužinu
![\overline{CD}](/media/m/3/3/8/338870e40f3ea7992d83158230115a5f.png)
. Ako sa
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
označimo polovište dužine
![\overline{CD}](/media/m/3/3/8/338870e40f3ea7992d83158230115a5f.png)
, odredite veličinu kuta
![\sphericalangle AMB](/media/m/2/6/2/262b2204f7c42814bb77c5bb097c91bc.png)
.
%V0
U paralelogramu $ABCD$ simetrala kuta $\sphericalangle DAB$ raspolavlja dužinu $\overline{CD}$. Ako sa $M$ označimo polovište dužine $\overline{CD}$, odredite veličinu kuta $\sphericalangle AMB$.
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Neka je
![\angle DAB=\alpha](/media/m/1/c/f/1cf68bda4e7f6e1db44b295246d4434c.png)
. Tada je
![\angle MAD=\alpha /2](/media/m/d/f/f/dffbdd4eacd3d755a366dedfaa6e49d9.png)
,
![\angle MDA=180-\alpha](/media/m/9/2/1/921eeeab7789f176f3a7d88599567ec8.png)
,
![\angle DMA= \alpha/2](/media/m/e/6/1/e6198a1cb6aa76cb6092f1406f7ff631.png)
,
![CD/2=AD=DM=MC=BC](/media/m/c/e/8/ce84fa10c34ae45a7c4060a9dd5d232c.png)
,
![\angle CMB=90- \alpha/2=\angle CBM](/media/m/f/6/f/f6f0af2b46f77ed872e3e108cc52dc05.png)
zbog
![CM=CB](/media/m/a/4/1/a412655d7578eaaa19ca19281ab2fe9c.png)
i
![\angle DB=\alpha](/media/m/b/5/6/b56affbb78812899b13323ccdefbd0c0.png)
,
![\angle AMB= 180- \alpha/2 - (90 - \alpha/2)=90](/media/m/7/9/6/7962fe56886786c8dc5ec32fc849ef7e.png)
.
%V0
Neka je $\angle DAB=\alpha$. Tada je $\angle MAD=\alpha /2$, $\angle MDA=180-\alpha$, $\angle DMA= \alpha/2$, $CD/2=AD=DM=MC=BC$, $\angle CMB=90- \alpha/2=\angle CBM$ zbog $CM=CB$ i $\angle DB=\alpha$, $\angle AMB= 180- \alpha/2 - (90 - \alpha/2)=90$.
13. siječnja 2016. 19:16 | ikicic | Točno |
14. siječnja 2016. 19:19 | grga | Točno |