Question

In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC 45°, Find ∠BCD.​

Answer 1

∠BCD = 80°

Explanation:

Given -

ABCD is a cyclic Quadrilateral.

∠DBC = 55°

∠BAC = 45°

To find -

∠BDC

Solution..

• BC is a segment and angles of same segment are equal.

=> ∠BAC = ∠BDC

=> 45° = ∠BAC { Angles of same segment BC}

In Triangle BCD

=> ∠BDC + ∠DBS + ∠BCD = 180°

=> 45° + 55° + ∠BCD = 180°

=> 100° + ∠BDC = 180°

=> ∠BDC = 180-100

=> 80° ans.

Answer 2

Given,

ABCD is a cyclic quadrilateral in which AC and BD are its diagonals.

`\: \: \: \: \: \: \: \: \: \: \: \:`

`\sf∠DBC= 55° and ∠BAC = 45°`

`\: \: \: \: \: \: \: \: \: \: \: \:`

`\sf∠CAD = ∠DBC= 55° (Angles \: in \: the \: same \: segment)`

`\: \: \: \: \: \: \: \: \: \: \: \:`

Thus,

`\sf∠DAB=∠CAD + ∠BAC`

`\: \: \: \: \: \: \: \: \: \: \: \:`

`\: \: \: \: \: \: \: \: \: \: \: \: \: \sf= 55° + 45°`

`\: \: \: \: \: \: \: \: \: \: \: \:`

`\: \: \: \: \: \: \: \: \: \: \: \: \sf= 100°`

`\: \: \: \: \: \: \: \: \: \: \: \:`

We know that the opposite angles of a cyclic quadrilateral are supplimentary.

`\: \: \: \: \: \: \: \: \: \: \: \:`

`\sf ∠DAB +∠BCD = 180°`

`\: \: \: \: \: \: \: \: \: \: \: \:`

`\sf∠BCD = 180°-100° =80°`

`\: \: \: \: \: \: \: \: \: \: \: \:`

`\pink{\boxed{\sf{∠BCD = 80°}}}`