The measures of the angles A, B, C of a triangle ABC are inversely proportional to the numbers: 0, (3), 0.5, 0.25. A parallel to the side BC intersects the perpendicular from C to point D. Find out the measures angles in the quadrilateral ABCD.
RE: Angles of a triangle, proportional
We have this image
And now this string of raports
and now it becomes
and find out the angles of the ABC triangle
m(A) = 60°
m(B) = 40°
m(C) = 80°
Now we have to do m(BAD) = m(BAC) + m(CAD) = m(BAC) + m(ACB) = 60° + 80° = 140°
m(CAD) = m(ACB) because they are two angles formed by two parallel lines cut by a secant.
And then, in the quadrilateral ABCD, we have m(B) = 40° m(BCD) = 90° m(D) = 90°
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