Question

in the given figure. O is the centre of the circle angle BCO=30° find X and y​

Answer 1

X and Y equals 30° & 15° respectively

Step-by-step explanation:

• The angle subtended by arc circle at the center = 2 \times the angle subtended by it on remaining part

Arc CD subtends $\angle COD$ at center and subtends $\angle BCD$ at B on the circle

Hence $\angle COD = 2\angle BCD$

That is $\angle COD = 2y\ [Since \angle BCD = y]$

Also $\angle COD = \angle OCB$= 30° [Alternate angles]

That is 2y = 30°

• Therefore, y = 15°

From the figure $\angle AOD$ = 90° since $\angle AEB$ = $\angle AEC$ = 90°

Therefore, $\angle AOD = 2\angle ABD$

That is 90° = 2 ∠ABD

Hence $\angle ABD = 45$°

In $\Delta AEB, \angle AEB$ + x + y + 45° = 180°

90°+ x + 15°+ 45° = 180°

• x = 180° - 150° = 30°

Answer 2

Answer:

hope this helps you :)

Step-by-step explanation: