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```
ABCD is a cyclic trapezium where AB||CD. if AB=4, AD=5, BD=7, find CD

```
9 years ago

```							The answer is CD=6.
Solution: draw BE parallel to AD such that BE=5 and DE=4. Let CE=x.
then, let angle BEC=Z
so, angle ADE and angle BCD are also Z(corresponding angles and cyclic trapezium).
From triangles BED and BEC,
cos 180-z=-cos z=(4^2+5^2-7^2)/2.4.5=-(5^2+x^2-5^2)/2.5.x
triangle BED                   Triangle BEC
on solving we get x=2 . so, CD=4+X=6.
```
9 years ago
```							thankyou so much!!!

but i didnt get why you took
cos z=(4^2+5^2-7^2)/2.4.5=-(5^2+x^2-5^2)/2.5.x
triangle BED                   Triangle BEC

i think basically i donno cos funcn in obtuse triangle
```
9 years ago
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