In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that `/_B E C\ =\ 130o` and `/_E C D\ =\ 20odot` Find `/_B A Cdot`

Explanation

Please refer to video for the figure.

Here, `/_BEC = 130^@ and /_ECD = 20^@`
`/_AEB+/_BEC = 180^@`(Sum of angles formed by a straight line is 180^@)

`/_AEB = 180-130 =50^@`

Also, `/_ABD = /_ACD` as these angles are subtended by a same chord on the circle.

So,` /_ABD = /_ABE = 20^@`

Now, in `Delta AEB`

`/_ABE+/_AEB+/_EAB = 180^@`

...