# Angle sums of a triangle

Given only the external angles of a triangle circumscribed in an arc and given the angle at the arc's centre that is between the vertices of the triangle's base. solve the triangle if the base of this triangle is a chord to the circle of which the arc is a part.solve the inscribed triangle

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Question giving orders? 0% accept rate? Two bad signs. –  Did Jun 24 '11 at 8:58
Could you provide a link to this problem, if there's any? –  Américo Tavares Jun 24 '11 at 11:11
What does "solve the triangle" mean? Surely you can't find the sides - if you scale all the lengths by any non-zero number, the angles stay the same. –  Gerry Myerson Jun 24 '11 at 12:55
@Didier: Third bad sign: he hasn't answer any comments in nay of his questions –  leonbloy Jun 24 '11 at 13:58
@leonbloy: Indeed. –  Did Jun 29 '11 at 14:16

## 1 Answer

As pointed out in the comments, you have not said that any length information was given, so it will not be possible to find any lengths, since the entire figure can be scaled (dilated) without changing any angles. If you know the external angles of a triangle, you can find the internal angles, since internal and external angles of a convex polygon form linear pairs (they are supplementary and their measures sum to 180°), without any of the stuff about the circular arc.

If you know the radius of the circular arc, then it's possible to find the length of the chord for which you know the subtended arc measure (if the radius is $r$ and the arc measure is $\alpha$, then the Law of Cosines applied to the triangle formed by the central angle and the chord says that the chord length $c$ satisfies $c^2=2r^2-2r^2\cos\alpha$). With one side length and all three interior angle measures, you can find a second side length by applying the Law of Sines, then use the Law of Cosines (or the Law of Sines) to find the third side length.

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