Question

please answer my question

Answer 1

CDB=180°-104°=76°

DBC=180°-(CDB+ACB)

=180°-(76°+65°)

=180°- 141°

=39°

IN ∆ ABD

A=3DBA

B=DBA

D=104°

A+B+D=180°

3DBA+1DBA+104°=180°

4DBA=180°-104°

4DBA=76°

DBA=76°/4

DBA=19°

A=3DBA

=3×19°

=57°

IN ∆ABC,

A=57°

C=65°

B=?

A+B+C=180°

57°+B+65°=180°

B+122°=180°

B=180°-122°

B=58°

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