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Given:

$A = (0,0)$

$B = (0,-10)$

$AB = AC$

Using the angle between $AB$ and $AC$, how are the coordinates at C calculated?

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    Aha, missed those four cases. Thank you Isaac!2010-08-23

3 Answers 3

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edit (to match revised question): Given your revised question, there is still the issue of C being on either side of the y-axis, but you have specified that AB=AC and that you are given $\mathrm{m}\angle BAC$ (the angle between AB and AC), so as in my original answer (below), the directed (trigonometric) measure of the angle from the positive x-axis to AC is $\mathrm{m}\angle BAC-90^{\circ}$ and AC=AB=10, so C has coordinates $(10\cos(\mathrm{m}\angle BAC-90^{\circ}),10\sin(\mathrm{m}\angle BAC-90^{\circ}))$. (This matches up to one of the answers in Moron's solution; the other corresponds to the other side of the y-axis.)

original answer (when it was not specified that AB=AC and when the given angle was C): As suggested in the comments, there are several cases. First, C could be on either side of the y-axis; let's assume that C has positive x-coordinate (leaving the case where it has negative x-coordinate for you to solve).

Second, ABC could be isosceles with AB=AC, AB=BC, or AC=BC. In the first case, $\angle B\cong \angle C$ (which cannot happen unless C is acute) and $\mathrm{m}\angle BAC=180^{\circ}-2\mathrm{m}\angle C$, so the directed (trigonometric) measure of the angle from the positive x-axis to AC is $90^{\circ}-2\mathrm{m}\angle C$ and AC=AB=10, so C has coordinates $(10\cos(90^{\circ}-2\mathrm{m}\angle C),10\sin(90^{\circ}-2\mathrm{m}\angle C))$. The second case is similar to the first (so it's left for you to solve). In the third case, C is equidistant from A and B, so C must lie on the perpendicular bisector of AB (as in J. Mangaldan's comment), and by symmetry this perpendicular bisector of AB also bisects $\angle ACB$; from there, you can use right triangle trigonometry to determine the coordinates of C (left for you to solve).

The cases where AB=AC (blue), AB=BC (red), and AC=BC (green) (lighter versions on the left side of the y-axis) are shown below for measures of angle C between 0 and 180°.

animated diagram

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    Sorry, I was bei$n$g stupid a$n$d using degrees when I was supposed to be using radians.2010-08-24
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Use Polar Co-ordinates. A point with polar co-ordinates $(r,\theta)$ is the same point in $x,y$ co-ordinates (or as also called, rectangular co-ordinates) as $(r\cos \theta, r\sin \theta)$.

In this case, point $C$ lies at a distance $10$ from $A$ which is the origin, so $r = 10$.

If given angle CAB is $\alpha$, the the polar angle is either $\frac{3pi}{2}-\alpha$ or $\frac{3pi}{2}+\alpha$, i.e in polar co-ordinates $C$ is either $(10,\frac{3pi}{2}-\alpha)$ or $(10,\frac{3pi}{2}+\alpha)$.

(It might help to draw a figure).

Now convert back to $x,y$ co-ordinates.

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    @Moron: $a$h, yes, $a$s J. Mangaldan indicated, I'd missed the edit.2010-08-23
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Let $a,b$ and $c$ be the side lengths and $A,B$ and $C$ the angles.

$a^{2}=x^{2}+\left( y+10\right) ^{2}$

$b^{2}=x^{2}+y^{2}=10^{2}$

$b=c=10$

By the (Neper) theorem of tangents (corollary of the Law of tangents):

$\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}$

On the other hand

$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C<\pi $

and by the theorem of sinus

$c\sin A=a\sin C\iff \left( x^{2}+\left( y+10\right) ^{2}\right) \sin C=10\sin A$

Compiling, we get:

$\frac{A-B}{2}=\arctan (\frac{\sqrt{x^{2}+\left( y+10\right) ^{2}}-10}{\sqrt{% x^{2}+\left( y+10\right) ^{2}}+10}\cot \frac{C}{2})$

$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C<\pi $

$(x^{2}+(\sqrt{100-x^{2}}+10)^{2})\sin C=10\sin A$

$y^{2}=10^{2}-x^{2}$

We have to solve this system of four equations and four unknowns $x,y,A,B$.

Edit: I started this approach before the question has been updated.

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    My first or second year teaching, a student who'd just learned about the Laws of Sines and Cosines asked if there was a Law of Tangents, so I pulled it out of an old [Dociani](http://en.wikipedia.org/wiki/Mary_P._Dolciani) book; something like 5 years later, I was watching a timed competition (state final) in which one contestant got the answer in about 30 seconds and no one else did because the question was a rearrangement of the Law of Tangents and only the one contestant knew it (none of my fellow coaches knew how she got it that fast).2010-08-23