The parallel sides of a trapezium have lenghts 5cm and 8cm..The other two sides have the lenghts 6cm and 4cm..Find the angles of the trapezium?

Answer 1

I constructed my trap ABCD, so that
AB=5, BC=4, CD=8, and DA=6
with BC and AD both having a negative slope, (leaning to the left)
I dropped perpendiculars from B and from A to meet CD at M and N respecively.
(M is on the extension of DC)
I let MC = x, then ND = 3+x
Notice that BNMA is a rectangle.

let the height BM = AN = h

two right-angled triangles:
x^2 + h^2 = 16 --> h^2 = 16-x^2
and
(3+x)^2 + h^2 = 36 --> h^2 = 36-(3+x)^2

16-x^2 = 36-(3+x)^2
16-x^2 = 36 - 9 - 6x - x^2
6x = 11
x = 11/6

then h^2 = 16 - 121/36 = 455/36
h = √455/6

now use basic trig ratios to find any angle you want
e.g. cos(angle BCM) = x/4
= (11/6)/4 = 11/24
angle BCM = 62.72°
so angle BCD = 180-62.72 = 117.28°
and angle ABC = 62.72°

Repeat for the other side of the trapezoid

Answer 2

Thanks!