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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
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Other answers:

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

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