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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`

Answer» 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^@`
`EAB+50+20 = 180`
`EAB= 110^@`
So, `/_BAC = EAB = 110^@`


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