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Given,

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

[tex] \: \: \: \: \: \: \: \: \: \: \: \:[/tex]

[tex] \sf∠DBC= 55° and ∠BAC = 45°[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \:[/tex]

[tex] \sf∠CAD = ∠DBC= 55° (Angles \: in \: the \: same \: segment)[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \:[/tex]

Thus,

[tex] \sf∠DAB=∠CAD + ∠BAC[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \:[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \sf= 55° + 45° [/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \:[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \sf= 100°[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \:[/tex]

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

[tex] \: \: \: \: \: \: \: \: \: \: \: \:[/tex]

[tex] \sf ∠DAB +∠BCD = 180°[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \:[/tex]

[tex] \sf∠BCD = 180°-100° =80°[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \:[/tex]

[tex] \pink{\boxed{\sf{∠BCD = 80°}}}[/tex]

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Answer:

∠BCD = 80°

Step-by-step 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.

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