Forum - Geometry

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
triunghi unghiuri invers proportionale
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 }}{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°
 
 

Search


Forum

Here you can discuss about mathematic, about algebra, geometry, trigonometry.

It is not mandatory to be logged in on this forum but it is nice to have an account. You can ask about mathematics just with your name and your email.

This maths forum is one of the easiest forums to use it.

>> Go to Math Forum

Forum Maths