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Answer 1

QUESTION

In the given figure. O is the centre of the circle angle BCO=30° find X/Y

Answer:

X/Y=30/15=2/1

solution

⇒X and Y equals 30° & 15° respectively

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

∴Arc CD subtends  at center ∠cod and subtends ∠bcd  at B on the circle

⇒Hence  ∠cod=∠bcd

That is  ∠cod=2y

Also ∠cod =∠ocb = 30° [Alternate angles]

⇒That is 2y = 30°

⇒therefore, y = 15°

⇒From the figure ∠aod = 90° since ∠aeb =∠aec  = 90°

⇒Therefore,  ∠aod=2∠aeb

• That is 90° = 2 ∠ABD

⇒Hence ∠abd=45°

⇒In triangle aeb ,∠aeb+ x + y + 45° = 180°

⇒90°+ x + 15°+ 45° = 180°

⇒x = 180° - 150° = 30°

X/Y=30/15=2/1

HOPE THIS HELPS U!

Answer 2

Answer:

Ok so in this, ABDC IS A CYCLIC QUADRILATERAL.

IN A CYCLIC QUADRILATERAL, THE OPPOSITE ANGLES ARE SUPPLEMENTARY. (180 DEGREES)

SO ANGLE A+ ANGLE D (y) IS 180 DEGREES

SO, 60+y= 180

y= 180-60

y= 120 DEGREES

ANGLE O= 2 times angle A

since angle A= 60 DEGREES

THEN ANGLE O WILL BE 60x2= 120 degrees

angle OBC AND ANGLE OCB ARE QUAL

THEN 180-120=60 DEGREES

60 DIVIDED BY 2= 30 degrees

so x=30 degrees

so y/x=

120 divided by 30

equals to 4

SO x/y = 4