Theres a triangle and the lines keep going, and I have to find the sum of the measures of the marked angles, which are vertical too the verticies
2007-11-29 08:54:43 · 5 answers · asked by Konstantin 3 in Science & Mathematics ➔ Mathematics
The triangle is equilateral, but the angles that I have to find , don't look like the same degrees..one is more obtuse than the other
2007-11-29 09:44:08 · update #1
If all of the angles marked are vertical to the interior angles (which is how I interpret the question), then the sum is 180 degrees. This follows from the fact that vertical angles are equal and interior angles of a triangle sum to 180 degrees.
It doesn't matter what the angles are, since you know that each angle is the same as the one vertical to it.
I'm assuming by vertical you mean "across from the intersection".
2007-11-29 09:16:49 · answer #1 · answered by ultimatelyconfused 2 · 0⤊ 0⤋
Well, if the triangle is equilateral, then the formula will be
((3-2)180)/3
or 60 degrees for each angle.
If it is a right (isosceles) triangle, then the other two angles
other then the right angle will be 45 degrees (sum of the
angle measures of a triangle is always 180 degrees).
Other then that, I can't help you unless the side lengths are
given.
2007-11-29 17:00:10 · answer #2 · answered by ozuyatamutsu 3 · 0⤊ 0⤋
Law of cosines maybe?
c^2 = a^2 + b^2 - 2ab·cos(angle)
Solve for the angle:
Angle = arccos[(c^2-a^2-b^2)/2ab]
c is the side opposite the angle you are trying to find, and a and b are the other two sides (doesn't matter which).
2007-11-29 17:02:48 · answer #3 · answered by someone2841 3 · 0⤊ 0⤋
They would be 120 degrees
2007-11-29 18:37:45 · answer #4 · answered by Anonymous · 0⤊ 0⤋
none of the angles are labeled at all?
2007-11-29 16:57:03 · answer #5 · answered by Anonymous · 0⤊ 0⤋