Given a trapezoid ABCD Find CD and AC?

0 ⤊

Given a trapezoid ABCD Find CD and AC?

Sketch: http://img68.imageshack.us/img68/4919/trapezoidle6.jpg

2006-11-28 06:03:24 · 3 answers · asked by James S 1 in Science & Mathematics ➔ Mathematics

3 answers

Assuming the the angles A and B are 90°...

Drop the vertical from vertex C. You now have a right triangle CDE (e is the point on AD where the vertical lands), height=3 and theta=48°. So CD=3/sin48°.

Now BC - AD - ED = 15 - 3/tan48°.

So AC = sqrt(AB^2 + BC^2) = sqrt(9 + (15 - 3/tan48°)^2).

2006-11-28 06:24:41 · answer #1 · answered by Anonymous · 0⤊ 0⤋

Tan 48 = 3/x, solve for x

CD^2 = 3^2+x^2 or CD=3/sin 48

3^2+(15-x)^2 = AC^2

2006-11-28 14:25:50 · answer #2 · answered by modulo_function 7 · 0⤊ 0⤋

that takes to long

2006-11-28 14:08:22 · answer #3 · answered by Aaron C 1 · 0⤊ 0⤋