troubling with trigonometry

[photon]

this evening i found a problem with trigonometry....of course it's just a Q of mine,not really a problem to you.
we learn trigonometry with right angled triangle in text book,but it can be also applied for acute and obtuse triangle(i guess).now if i want to get sine or cosine of any angle ,i need hipotenuse,adjacent,opposite.....how to find them?ok,we can get opposite as the oppposite side of the angle(we have taken).but,what about hipotenuse and adjacent?
if everything is right stated above about using trigonometry on obtuse and acute triangle,i got an idea about hipo and adjacent.certainly, hipotenuse>adjacent......
so the greater of the SONNIHITO sides of the angle will be so called hipo and the other (smaller) will be so called adjacent.
am i right????

[tarek like math]

i think u don't understand trigonometric ratio, you can read my solution below.to measure trigonomitric ratio of an angle you should first select a point of one arms . your selected arm must not base means bhumi . from that point you must draw a line to base which create right angle with base. suppose ab is base then bc is adjacent arm and your selected point is c in bc arm. now draw a right angle <cda. now hypotenuse means 'atibhuj' is ac, base mean bhumi is ad and 'Lambo' is cd. so now you can measure many trigonometric ratio. for example sin(cad)=cd/ac, cos(cad)=ad/ac. these ratio will not change if you change your point c.
you may guess that in right angle cd=ac=1

[tarek like math]

i think u don't understand trigonometric ratio. to measure trig: ratio of an angle you can read my note below.
suppose your angle is <cab. here ab is base mean 'bhumi'. ac is adjacent arm. now to measure trigonometric ratio u must select a point in ac arms . then u should draw a line from that point to ab which create right angle with ab. think that ur selected point is c in ac arm and u draw a line cd to ab which mean angle <cda is right angle. so now ur hypotenuse mean 'atibhuj' is ac , base mean 'bhumi' is ad and 'lambo' is cd. now u can measure all trigonometric ratio .
for example sin(cad)=cd/ac , cos(cad)=ad/ac. these ratios will not change if u change ur selectd point .
you may guess that in case of right angle ur hypotenuse and lambo are same.
mean ac=cd=1 this is how sin90=1
{from tarek }

[Moon]

Photon: I have moved the topic to secondary sub-forum. I think that it is not a Olympiad geometry problem. So you better posted it here in mathematics section fro secondary students.

@Tarek: Welcome to the forum. You can write nice equations in this forum. So please learn how to use LaTeX codes to write equations.

[photon]

i got an idea about hipo and adjacent.certainly, hipotenuse>adjacent......
so the greater of the SONNIHITO sides of the angle will be so called hipo and the other (smaller) will be so called adjacent.
am i right????

i sometimes get so much idiotic ideas.....
yes, it is not possible.thinking over it ,i got it,may be called proof for ''we need right angle for trigonometry''.
consider a triangle ABC where no angle is 90 degree,angle B is acute.AB>BC.according to my writing ''without right angle, we can use trigonometry''.so,$tanB=AC/BC$(just applying my wrong idea)
we take some points P,Q,R from the extention of AB behind B point.add P,C and Q,C and R,C.the we can write
$tanB=CP/BC=CQ/BC=CR/BC$ according to triangles BCP,BCQ,BCR.( again apply)
we get,CP=CQ=CR.P,Q,R were taken randomly ,we can say every point in AB line has equal distance from C!!! not only these,but also
\[\angle CPQ=\angle CQP\]
\[\angle CQR=\angle CRQ\]
\[\angle CPQ=\angle CRQ\]
so,\[\angle CQP=\angle CQR\]
it means CP,CQ,CR are perpendicular to AB from C and we can get infinite perpendicular to AB from C!!!!
i applied my wrong idea considering it true but it is impossible.so,it not possible to apply trigonometry without right angle.
@tarek like math,i knew what you wrote,actually i was just confused about.....ulta palta jottosob chinta