# Find the coordinates in an isosceles triangle

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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The most obvious hint would be that C should lie in the perpendicular bisector of A and B. You should be able to reckon the needed equations from there. –  Ｊ. Ｍ. Aug 23 '10 at 13:53
Which two sides of the isosceles triangle are congruent? –  Isaac Aug 23 '10 at 13:54
Now that I think about it... if the only constraint for C is the angle corresponding to that corner, you have two solutions. –  Ｊ. Ｍ. Aug 23 '10 at 13:56
@J. Mangaldan: I think an acute angle C will give 6 solutions (2 distinct triangles, swapping A and B in one of those two triangles, symmetry over the y-axis), while a right or obtuse angle C will only give 2 solutions (symmetry over the y-axis). –  Isaac Aug 23 '10 at 14:00
@J. Mangaldan: Only if you are assuming AC=BC (as opposed to AB=AC or AB=BC) –  Isaac Aug 23 '10 at 14:10

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°.

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Given the update, we're back to considering two cases. –  Ｊ. Ｍ. Aug 23 '10 at 15:05
+1: Can you please remove the irrelevant portion (or move it to the end, rather than keep it at the beginning?) –  Aryabhata Aug 23 '10 at 16:09
Didn't want to waste the programming effort, 'no? ;) –  Ｊ. Ｍ. Aug 23 '10 at 16:57
Right. Sadly I cannot vote twice! –  Ｊ. Ｍ. Aug 23 '10 at 17:02
+1 for both the solution and the picture. –  Américo Tavares Aug 23 '10 at 17:36

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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This is only true in the case where AB=AC (which is not possible unless C is an acute angle). –  Isaac Aug 23 '10 at 15:01
@Isaac: The question states AB=AC, and that angle BAC is given. –  Aryabhata Aug 23 '10 at 15:01
Moron (I feel uncomfortable with your user name): the new info wasn't there when Isaac and me were playing around. –  Ｊ. Ｍ. Aug 23 '10 at 15:06
@J. Mangaldan: I see, I only added this answer to specifically mention polar co-ordinates. If Isaac edits his answer to match the question, I will surely upvote his answer. –  Aryabhata Aug 23 '10 at 15:11
I upvoted yours, since the situation the OP now presents is indeed neatly done in polar coordinates (i.e. very convenient that the apex of the isosceles triangle is at the origin). –  Ｊ. Ｍ. Aug 23 '10 at 15:14

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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Heh, I haven't seen the name Neper in quite a while! –  Ｊ. Ｍ. Aug 23 '10 at 15:40
@Mangaldan: I dont't know whether this theorem is known now as Neper's Theorem. This name is from my portuguese trigonometry book from the 1960's. It may be derived from the law of the sinus. –  Américo Tavares Aug 23 '10 at 15:46
I think that's a rearrangement of what I've seen called the Law of Tangents. –  Isaac Aug 23 '10 at 15:59
@Isaac: I searched now at the wikipedia and added the link and saw it is a corollary of the Law of Tangents. –  Américo Tavares Aug 23 '10 at 16:16
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 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). –  Isaac Aug 23 '10 at 17:05