Question

In the given figure. O is the centre of circle,angle BCO=30 angle AEB=90 and OD || BC find x and y.

Answer 1

There are many ways to solve this. This is a simple answer. See the diagram.

OD || BC, OC is a transversal. Alternate angles: ∠DOA =∠OCB = 30°.

 

Arc CD makes ∠y on the circle & 30° at the center.  So y = 30°/2 = 15°.

 

OD || BC, AE  |  BC  =>  AE  |  OD ,  ∠AOD = 90°

Arc AD makes ∠ 90° at center. So it makes on the circle at B: ∠ABD = 45°.

 

∠ABE = 45°+15°=60°.

In the right angle ΔABE :   x = 90° - 60° = 30°.

Answer 2

Step-by-step explanation:

Value of X = 30°. And,

Value of Y= 15° .