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Math Question... Is It Possible?


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Posted

I said that when you consider it as separate triangles with the y as heights which would give x=t which doesnt make any sense thats what I was trying to tell..you chose y(small triangle) side as the base and then applied the similar triangle..thats the trick..
[quote name='jamajacha' timestamp='1352784962' post='1302788373']
thanks sleeping now..good night

if only x=t is the answer...there is point in solving this problem..
by observation of problem itself..... u can say
[/quote]

Posted

t can be any value on x, t is arbitrary and value of t can be found.. if the angle is known..

Posted

x= 3 . y = 2


t =1 kaniti ssooputho calculate sesa nia seppu.. evadina el ani adigithey ..

Posted

Inthaki correct answer enti?

Posted

[quote name='cavaliers' timestamp='1352736066' post='1302783125']
Guys,

I have the coordinates X and Y. Is there a way we can find t?



Thanks.

[img]http://i48.tinypic.com/6nuohi.jpg[/img]
[/quote]


bhayya without another variable, i think its not possible to find the value of t.
another variable may be the angle of another side in the triangle with side 't'. (!(<

Posted

[quote name='Aamphat' timestamp='1352815630' post='1302788844']
Inthaki correct answer enti?
[/quote]


bhayya na email

Posted

[quote name='jamajacha' timestamp='1352780493' post='1302788117']
[img]http://i47.tinypic.com/mkyu6w.jpg[/img]
[/quote]


[img]http://i48.tinypic.com/8x7eko.jpg[/img]


Thanks Bhayya,....

I guess I found I was looking for.. actually I have to find out where did the sensor hit on that object of 5 Mts... Looking at your images.. I thought Instead of looking from the start of the object I can also look from the end of the object.

so, the new t = (Y^2)/X solves the problem


Thanks again....

:)

Posted

[quote name='cavaliers' timestamp='1352822291' post='1302789303']
[img]http://i48.tinypic.com/8x7eko.jpg[/img]


Thanks Bhayya,....

I guess I found I was looking for.. actually I have to find out where did the sensor hit on that object of 5 Mts... Looking at your images.. I thought Instead of looking from the start of the object I can also look from the end of the object.

so, the new t = (Y^2)/X solves the problem


Thanks again....

:)
[/quote]
two traingles similar ga unaayyee antey chaalu...u can solve for any variable available..
the concept of trigonometry and properties of triangles has been developed the similar triangles concept

Posted

[quote name='CASANOVA' timestamp='1352821984' post='1302789277']


bhayya na email
[/quote]
ratri ki sending

Posted

[quote name='Aamphat' timestamp='1352829248' post='1302790035']
ratri ki sending
[/quote]

will wait

Posted

[quote name='cavaliers' timestamp='1352822291' post='1302789303']
[img]http://i48.tinypic.com/8x7eko.jpg[/img]


Thanks Bhayya,....

I guess I found I was looking for.. actually I have to find out where did the sensor hit on that object of 5 Mts... Looking at your images.. I thought Instead of looking from the start of the object I can also look from the end of the object.

so, the new t = (Y^2)/X solves the problem


Thanks again....

:)
[/quote]

idhi endi sudden ga bomma marindi..nee thesis title change chesava endi?

Posted

[quote name='CASANOVA' timestamp='1352831571' post='1302790255']
idhi endi sudden ga bomma marindi..nee thesis title change chesava endi?
[/quote]


Leedu bhayya.... Looking at the same image in a different prospective anthe...

Posted

[quote name='LOGON' timestamp='1352786430' post='1302788441']
t can be any value on x, t is arbitrary and value of t can be found.. if the angle is known..
[/quote]
[quote name='MAN UTD' timestamp='1352821728' post='1302789257']


bhayya without another variable, i think its not possible to find the value of t.
another variable may be the angle of another side in the triangle with side 't'. (!(<
[/quote]

Agreed...

Posted

[quote name='cavaliers' timestamp='1352822291' post='1302789303']
[img]http://i48.tinypic.com/8x7eko.jpg[/img]


Thanks Bhayya,....

I guess I found I was looking for.. actually I have to find out where did the sensor hit on that object of 5 Mts... Looking at your images.. I thought Instead of looking from the start of the object I can also look from the end of the object.

so, the new t = (Y^2)/X solves the problem


Thanks again....

:)
[/quote]


avi similar triangles etla ainayi dude.. avi similar triangles kaadu...

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