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Hmm to make it clearer, AB = AC
can be anything between 0 and 90
need more info, i guess
I assume there is an exact value ?! Note: that's all i've got for the question
analytically, whatever angle be BAC, above figure can be drawn. how can u find exact value??
agree with Arnab, I don't think there's enough info... Is BC given?
if the length of segment was given, i guess it would be possible
Sorry but if i knew, i wouldn't have asked. It's a question from my friend and that's all she gave me for the question. She said the answer was like 67.5
that's the answer for what?
length of BC?
@dpaInc angle ABC or ACB whatever,
AD=CD, it would be given
@Arnab09 it's not stated in the question. She read it from the answer
oh! sorry, @Callisto i guess not enough data given here
No problem, thanks for taking a look!
well i arrived at the answer your friend did
was your friend's answer right?
I hope so, she read from the solution
ok, then we must assume that AD=DC
but how would you make such assumption if you don't know the answer?
not sure on that one, maybe there is a theorem that states it
law of sines perhaps?
i dont think so ... third side is missing
if CD is given, the required angle can be calculated easily
If CD had been given, I wouldn't have asked
I think it can have any value between 45 to 90 ... since you can always draw perpendicular from one base angle to other side
i think descartes might work
Can you explain a little bit?
i cant draw on this thing for nothing; descartes would label all the lines; known and unknown; then write up all the relations he could think of until he could construct a set of n equations in n unknowns and solve
You don't have enough information. First triangle 1 equation two unknowns second triangle 1 equation 3 unknowns. Combine unknown angle from both equations and you have 1 equation two unknowns.
Okay, that's what I'm doing right now :S
you have pythag and labels and your own determination :)
the angle of A shouldnt care thata = 139; so make it easier for yourself; a = 1 is just as good
where is the theta?
up top, angle A is going to be a ratio of sides; so focus on sides
so far, i've got e^2 = 2d ...
a = 1 b+d = 1 ; d= 1-b b^2 + c^2 = 1; c^2 = 1-b^2 d^2 + c^2 = e^2 (1-b)^2 + (1-b^2) = e^2 1 -2b + b^2 +1 - b^2 = e^2 2-2b = e^2 2(1-b) = e^2 ; (1-b) = d 2d = e^2 me too
a^2 = b^2 +c^2 e^2 = c^2 + d^2 and a = b+d = 198 e^2 = c^2 + d^2 e^2 = a^2 - b^2 + d^2 = 198^2 - (d-b)(d+b) = 198^2 - [d-(198-d)] (198) = 198 ^2 + 2d - 198^2 e^2 = 2d
c^2 = e^2 - d^2 c^2 = 2d - d^2 c^2 = d(2-d) ; d = 1-b c^2 = (1-b)(2-(1-b)) c^2 = (1-b)(1-b) c^2 = (1-b)^2 c^2 + b^2 = 1 (1-b)^2 + b^2 = 1 1 -2b +b^2 +b^2 = 1 2b^2 -2b +1 = 1 2b(b-1) = 0 ; when b=0 or b=1 well i think ive narrowed it down to the obvious :) b can range from 0 to 1 lol
the range is too large.... :S it's like from 0 - 198 ....
yeah. 0 to a :)
|dw:1333435577728:dw| i think one parameter is missing
if b = a then A = 0 if b=0 then A = 0
could both case happen?
hmm, if b - 0, A = 90
is it possible to have 2 right angle in a triangle (I don't think so..)
But it's no longer an isos. triangle
the right side is a rt iso tri; the left is just a degenerate
a=a, d=a, b=0, c=a, e=d sqt(2), A = 90 if a=a, b=a, e=0, d=0, c=0, A = 0
Sorry... I don't understand.....
It simply means that A can range from 0 to 90; at zero you get a line; at 90 you get an iso rt triangle
as determined at the start, this thing really needs another piece of information to make A stand still and have a solid value
or ideally A = 60 degrees
Alright .. thanks ...
i did have fun trying to think like descartes tho ;)
he says, when you do all that you can do; thats all that can be done until you know more; so as is, we need to know more to make A stand still
Finally, my friend solved her own problem :S Thank you for all your effort!!!!!