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how do u use trigonometry to determine an angle of an isosceles triangle??

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Supposedly we are given the sides of isosceles triangle. If we draw a perpendicular from the vertex to the third side (two other sides are equal). It'll bisect the third side. |dw:1331409664200:dw|
We need to find the two angles x and the third angle y we know \[ \cos \theta= \frac{base}{hypotenuse}\] here \[\cos x= \frac{b/2}{a}\] from this we can find x and y=180-2x
angles are determined by undoing a ratio

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Other answers:

trig(angle) = ratio angle = arctrig(ratio)
which angle are you interested in finding?
both of them
there are 3 angles in a triangle :) I assume you mean the base angles?
yup! sorry bout that
in order to determine the solution to any triangle we need a few bits of information to deduce the unknowns with
do you have a particular example we can work on? :( its a studyguide for finals and i'm reviewing
well, there are 2 "laws" that can be used when we have certain information; one is the law of sines and the other is a more adaptable form of the pythag thrm
law of sines equates angles with the length of the side opposite the angle \[\frac{sin(C)}{c}=\frac{sin(B)}{b}=\frac{sin(C)}{c}\] |dw:1331413584401:dw|
the law of cosines is the more general form of the pythag thrm: \[c^2=a^2+b^2-2ab\ cos(C)\]
when the angle C is 90 degrees; that simply reverts the the usual: c^2=a^2+b^2
the law of cosines is good for determing the length of sides; and the law of sines is easier to implement once you know the ratio of angles and sides

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