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Question 258370: my question states to locate a position in triangle (SAB), that is equidistant from all the angles. Given that: SA is 8 cm, AB is 8 cm and SB is 11.4 cm. Angle S is 28 degrees, angle A is 90 degrees and angle B is 45 degrees.
Thank you for your help in advance.
billy-bob
Found 2 solutions by richwmiller, Theo:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! Angle s can't be 28 when angle b is 45
Since SA and AB both equal 8 cm, angle s and angle b must be the same size.
They can't both be 28 because that is too small.
Also the third side can't be 11.4 either. It must be 11.3 .
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! you want to find the point inside the triangle that is equidistance from all the vertex of the triangle.
I believe you want the circumcenter of the triangle.
this is the center of a circle that circumscribes the triangle.
as such, all vertices of the triangle are on the circumference of the circle which means they are the same distance to the center of the circle.
that point is equidistant from each vertex of the triangle.
here's a link that shows you what I mean.
http://www.mathopenref.com/trianglecircumcenter.html
that point is the intersection of the perpendicular bisectors of each side of the triangle.
that point can be inside or outside the triangle.
just play with the figure and you can see how that happens.
here's another look at it.
http://faculty.evansville.edu/ck6/tcenters/class/ccenter.html
in your triangle SAB, you have:
SA is 8 cm.
AB is 8 cm.
SB is 11.4 cm.
this should be an isosceles triangle since 2 sides of it are equal.
the legs are SA and AB and the base is SB.
you state that angle S is 28 degrees and angle A is 90 degrees and angle B is 45 degrees.
that can't be.
sum of the interior angles of a triangle is 180 degrees.
sum of the angles you provided is 163 degrees.
also, if the sides of a triangle are equal, then their corresponding angles are also equal.
This does not ring true with your statement about the sides and the angles.
something is wrong with your problem setup.
check it again and correct it and resubmit the problem.
in any case, if you want the point that is equidistance from the vertices of the triangle, then you are looking for the intersection of the perpendicular bisectors of each of the sides because that give you the center of the circle that circumscribes the triangle.
that point is equidistant from the vertices of the triangle.
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