# Angles of a triangle, proportional

Taylor Maria
Visitor
2016-06-21 18:53:30
#1
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

Stephan
Visitor
2016-06-21 19:01:01
#2
We have this image

And now this string of raports
$\frac{A}{\frac{1}{0,(3)}}=\frac{B}{\frac{1}{0,5}}=\frac{C}{\frac{1}{0,25}}$, i.e.
$\frac{A}{\frac{9}{3}}=\frac{B}{\frac{10}{5}}=\frac{C}{\frac{100}{25}}$ and now it becomes
$\frac{A}{3}=\frac{B}{2}=\frac{C}{4}=\frac{A+B+C}{3+2+4}=\frac{180^{\circ&space;}}{9}$
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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