Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



A different geometry problem
Posted:
Dec 10, 2010 4:49 AM


Hi,
It looks like it were a "Sujeet's" but it is a little different...
In a triangle ABC, a point P such as angle PAB = 30 deg, angle PAC = 20 deg, angle PBA = 10 deg, angle __PCB__ = 70 deg. Find the angle PBC. So that there is only one solution, I add the following condition : none of the triangles in the figure is isosceles.
This problem is different from the previous ones, as the given data results into _two_ solutions. One of them has some isosceles triangles and is then rejected by the additional condition.
It seems however you have to find both solutions to know which one is to be rejected !
Regards.



