Question

in the adjoining figure chord AC and BD of a circle with Centre O intersect at right angle at E if angle OAB equal to 25 degree calculate angle EBC

Answer 1

hi bhai
here's your answer looking for

construction :- join OB

now, We got AOB as iso. triangle , where OA = OB

so, angle OAB = angle OBA. ( opp. angle of iso. triangle )
Now, O B A = 25

so,
in triangle AOB ,

OAB + OBA + AOB = 180. ( angle sum property of triangle )

$25 + 25 + aob = 180 \\ \\ aob = 180 - 50 \\ \\ aob = 130$
now,

$\frac{1}{2} aob = acb$
( angle sustained by the same chord , the angle of centre will be 2 times of the angle on the circumference)

hence,

ACB = ECB = 1/2×130
=65

now, in triangle EBC,

CEB + EBC + ECB = 180

$90 + ebc + 65 = 180 \\ \\ ebc = 180 - 155 \\ \\ ebc = 25$
hope you are satisfied with my answer

Answer 2

Hey there,

In the adjoining figure chord, AC and BD of a circle with Centre O intersect at right angle at E if angle OAB which equal to 25.

Hope this helps!