Question

In fig . o and o' are centre of two circles intersecting at B and C .
ACD is a straight line , find x.

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Answer 1

Given:-

• ∠AOB = 130°
• ACD is a straight line.

To find:-

• ∠BO'D = x° = ?

Answer:-

In the circle with centre O,

We know that centre angle = 2 * circum angle

So,

∠AOB = 2 * ∠ACB

→ 130° = 2 * ∠ACB

→ ∠ACB = 65° ----( 1 )

In line segment ACD,

Since ACD is a straight line segment,

∠ACB + ∠DCB = 180° (linear pair)

→ 65° + ∠DCB = 180° [From ( 1 )]

→ ∠DCB = 180° - 65°

→ ∠DCB = 115° ----( 2 )

In the circle with centre O'

We know that centre angle = 2 * circum angle

So,

(reflex ∠BO'D) = 2 * ∠BCD

→ (reflex ∠BO'D) = 2 * 115 [From ( 2 )]

→ (reflex ∠BO'D) = 230°

In the circle with centre O'

(reflex ∠BO'D) + ∠BO'D = 360°

→ 230° + ∠BO'D = 360° [From ( 2 )]

→ ∠BO'D = 360° - 230°

→ ∠BO'D = 130°

So,

→ x° = 130° Ans.

Answer 2

Given : ACD is a straight line  ; ∠AOB = 130° ; ∠BO¹D = x°

∠AOB is an exterior angle for ΔOBC  ⇒ ∠AOB = ∠OCB + ∠OBC

also ΔOBC is an isosceles triangle with OC = OB = radius of circle

⇒ ∠OCB = ∠OBC

so, 130° = ∠OCB + ∠OBC = 2∠OCB

⇒  ∠OCB = 65°

Now, ∠ACB + ∠DCB = 180°  (∵ angle on a straight line ACD)

⇒ 65°  + ∠DCB = 180°

⇒ ∠DCB = 180° - 65°

⇒ ∠DCB = 115°

We know that angle at centre of a circle is twice the angle formed on the circumference

Here in circle with centre O¹, ∠BCD is the angle on circumference & reflex angle∠BO¹D is the angle at the centre

⇒ reflex∠BO¹D  = 2 x ∠BCD

⇒ reflex∠BO¹D = 2 x 115°

∴ reflex∠BO¹D = 230°

Now we know, reflex ∠BO¹D + ∠BO¹D  = 360°

 ⇒ 230° + x° = 360°

⇒  x° = 360° - 230°

     ∴ x° = 130°

HOPE THIS HELPS YOU !!   : )