Question

angle bco=30degree find x and y​

Answer 1

Answer:

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°.

hope it helps

Answer 2

Step-by-step explanation:

$\huge\underline\orange{Hola Mate,}$

Given O is the centre of the circle and ∠BCO = 30°.

Construction:

Join AC and OB.

In ΔOBC,

⇒ OC = OB [Radii of circle]

⇒∠OBC = ∠OCB = 30°

From angle sum property,

⇒∠OBC + ∠OCB + ∠BOC = 180°

⇒ 30° +30° + ∠BOC = 180°

⇒∠BOC = 180° - 60°

∴ ∠BOC = 120°

We know that the central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.

⇒∠BOC = 2∠BAC

⇒ 2∠BAC = 120°

∴ ∠BAC = 60°

⇒∠AEB = 90°

So, OE ⊥ BC

We know that a line form center to any chord is perpendicular then that line also bisects chord.

⇒ CE = BE … (1)

Consider ΔABE and ΔACE,

⇒ CE = BE [From (1)]

⇒∠AEB = ∠AEC = 90°

⇒ AE = AE [Common side]

By RHL rule,

⇒ ΔABE ≅ ΔACE

By CPCT,

∠BAE = ∠CAE

⇒∠BAC = ∠BAE + ∠CAE

⇒ x + x = 60°

⇒ 2x = 60°

∴ x = 30°

And ∠COI = ∠OCB = 30° [Alternate interior angles]

We know that the central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.

⇒∠COI = 2∠CBI

⇒ 2y = 30°

∴ y = 15°

Hope this helps you

Please make me as brainliest ✌️✌️✌️