highschoolmom2010
  • highschoolmom2010
The Polish astronomer Nicolaus Copernicus devised a method for determining the sizes of the orbits of planets farther from the sun than Earth. His method involved noting the number of days between the times that a planet was in the positions labeled A and B in the diagram. Using this time and the number of days in each planet’s year, he calculated c and d.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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highschoolmom2010
  • highschoolmom2010
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highschoolmom2010
  • highschoolmom2010
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highschoolmom2010
  • highschoolmom2010
@ganeshie8

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highschoolmom2010
  • highschoolmom2010
@thomaster
ganeshie8
  • ganeshie8
do u see a right triangle in that cosmic jungle ?
ganeshie8
  • ganeshie8
|dw:1376506909669:dw|
highschoolmom2010
  • highschoolmom2010
yes
ganeshie8
  • ganeshie8
lets solve question A first :- say, x is the distance between Sun and Mars
ganeshie8
  • ganeshie8
|dw:1376507051350:dw|
ganeshie8
  • ganeshie8
|dw:1376507103333:dw|
ganeshie8
  • ganeshie8
can u find that angle ? (look at the cosmic jungle pic again)
ganeshie8
  • ganeshie8
that angle equals d-c = 103.8-55.2 = ?
ganeshie8
  • ganeshie8
once u have that angle, u can find the x easily using an appropriate trig ratio
highschoolmom2010
  • highschoolmom2010
48.68
ganeshie8
  • ganeshie8
103.8 - 55.2 48.6
ganeshie8
  • ganeshie8
its just 48.6
ganeshie8
  • ganeshie8
how did u get 48.68 ?
highschoolmom2010
  • highschoolmom2010
103.88accidentally pushed the |dw:1376507391898:dw|
ganeshie8
  • ganeshie8
lol okie :)
highschoolmom2010
  • highschoolmom2010
so now we know that
ganeshie8
  • ganeshie8
|dw:1376507462089:dw|
ganeshie8
  • ganeshie8
u tell me, which ratio we can use here to find distance between Suna and Mars(x)
highschoolmom2010
  • highschoolmom2010
|dw:1376507504635:dw|
ganeshie8
  • ganeshie8
forget about million years. lets do the calculations in AUs which are simple
highschoolmom2010
  • highschoolmom2010
we know hyp and adj SOH CAH TOA we use cos
highschoolmom2010
  • highschoolmom2010
how do i do that
ganeshie8
  • ganeshie8
yes cos ! and pls use AUs...
highschoolmom2010
  • highschoolmom2010
how do i do that
ganeshie8
  • ganeshie8
and the question is also asking u for AUs oly
ganeshie8
  • ganeshie8
|dw:1376507639742:dw|
ganeshie8
  • ganeshie8
hyp = x adj = 1 AU
highschoolmom2010
  • highschoolmom2010
oh ok so i just put 1 AU not 93 million okie :D
ganeshie8
  • ganeshie8
AU is just an unit, like meter or second. so dont bother about it much :)
ganeshie8
  • ganeshie8
yes :)
ganeshie8
  • ganeshie8
cos 48.6 = 1/x solve x
ganeshie8
  • ganeshie8
do this, and do the Jupiter problem also same way. i need to go have dinner. brb.
highschoolmom2010
  • highschoolmom2010
\[\cos 48.6=\frac{ 1 }{ x }\] \[x *\cos 48.6=\frac{ 1 }{ ?x }*x\] \[x \cos 48.6=1\]
highschoolmom2010
  • highschoolmom2010
\[\frac{x \cos 46.8 }{ \cos 46.8 }=\frac{ 1 }{ \cos 46.8 }\] \[x=\frac{ 1 }{ \cos 46.8 }\]
highschoolmom2010
  • highschoolmom2010
do i do the inverse now
ganeshie8
  • ganeshie8
just plugin that \(\large x = \frac{1}{\cos 48.6}\)
ganeshie8
  • ganeshie8
you should get x = 1.5 AU approx.. try the Jupiter problem...
highschoolmom2010
  • highschoolmom2010
|dw:1376587258601:dw|
highschoolmom2010
  • highschoolmom2010
|dw:1376587399528:dw|
highschoolmom2010
  • highschoolmom2010
\[\cos 100.8=\frac{ 1 }{ x }\] \[(x)* \cos 100.8=\frac{ 1 }{ x }*x\] \[ X \cos 100.8=1\] \[\frac{ x \cos 100.8 }{ \cos 100.8 }=\frac{ 1 }{ \cos 100.8 }\]
ganeshie8
  • ganeshie8
angle is not 100.8
ganeshie8
  • ganeshie8
|dw:1376587971593:dw|
ganeshie8
  • ganeshie8
for jupitor c = 21.9, d = 100.8 angle = d-c = ?
highschoolmom2010
  • highschoolmom2010
oh... 78.9
ganeshie8
  • ganeshie8
yes do it again :)
highschoolmom2010
  • highschoolmom2010
\[\cos 78.9=\frac{ 1 }{ x }\] \[(x)* \cos 78.9=\frac{ 1 }{ x }*(x)\] \[x \cos 78.9=1\] \[\frac{ x \cos 78.9 }{\cos 78.9 }=\frac{ 1 }{ \cos 78.9 }\] \[x \approx 5.19\]
highschoolmom2010
  • highschoolmom2010
@ganeshie8
ganeshie8
  • ganeshie8
perfect !
ganeshie8
  • ganeshie8
now tell me this, did u get why we have been doing d-c for angle ?
highschoolmom2010
  • highschoolmom2010
no not really
ganeshie8
  • ganeshie8
look at the origianal cosmic pic, you wil understand
ganeshie8
  • ganeshie8
observe that, \(d\) is the angle moved by earth
ganeshie8
  • ganeshie8
and \(c\) is the anle movied by the other planet in same time
highschoolmom2010
  • highschoolmom2010
i see that
ganeshie8
  • ganeshie8
\(d-c\) gives u the angle in that right triangle (visually u need to **see** it)
ganeshie8
  • ganeshie8
|dw:1376589514631:dw|
ganeshie8
  • ganeshie8
|dw:1376589569683:dw|
ganeshie8
  • ganeshie8
that entire angle is \(d\)
highschoolmom2010
  • highschoolmom2010
|dw:1376589610421:dw| right ?
ganeshie8
  • ganeshie8
that part is \(c\)
ganeshie8
  • ganeshie8
yes, right !
ganeshie8
  • ganeshie8
So, whats the angle in that triangle ?
ganeshie8
  • ganeshie8
|dw:1376589698138:dw|
highschoolmom2010
  • highschoolmom2010
78.9
ganeshie8
  • ganeshie8
uhh i mean in terms of \(d\) and \(c\)
highschoolmom2010
  • highschoolmom2010
im not sure what u mean
ganeshie8
  • ganeshie8
|dw:1376589806634:dw|
highschoolmom2010
  • highschoolmom2010
oh ok
ganeshie8
  • ganeshie8
angle in the triangle equals \(d - c\) you worked the problem correctly ! just wanted u to see why we did \(100.8-21.9 \)
ganeshie8
  • ganeshie8
hope it makes sense, the problem is easy actually - except for the fancy copernicus and planets words..
highschoolmom2010
  • highschoolmom2010
yea the AU threw me off
ganeshie8
  • ganeshie8
haha i thought so :)
highschoolmom2010
  • highschoolmom2010
i think that was just uncalled for work
ganeshie8
  • ganeshie8
hmm -_-
highschoolmom2010
  • highschoolmom2010
ty @ganeshie8
ganeshie8
  • ganeshie8
np :)

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