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Answer:
[tex]( \angle BAC = 80^\circ )[/tex]
[tex]( \angle ABC = 60^\circ )[/tex]
[tex]( D ) \text{is a point such that:}[/tex]
[tex]( \angle DAB = 10^\circ )[/tex]
[tex]( \angle DBA = 20^\circ )[/tex]
[tex]2. \text{Determine} \: ( \angle ADB ):[/tex]
[tex]{ ( \angle DAB + \angle DBA + \angle ADB = 180^\circ ) }[/tex]
(the sum of angles in triangle ( ABD )),
[tex]( 10^\circ + 20^\circ + \angle ADB = 180^\circ ),[/tex]
[tex]( \angle ADB = 180^\circ - 30^\circ = 150^\circ ).[/tex]
[tex]3. \text{Determine} ( \angle BCA ) in triangle ( ABC ):[/tex]
The sum of angles in triangle ( ABC ):
[tex]{( \angle BAC + \angle ABC + \angle BCA = 180^\circ ),}[/tex]
[tex]( 80^\circ + 60^\circ + \angle BCA = 180^\circ ),[/tex]
[tex]( \angle BCA = 180^\circ - 140^\circ = 40^\circ ).[/tex]
4. Determine ( angle ACD ):
Since ( D ) lies inside triangle ( ABC ), ( angle ACD ) is an exterior angle for triangle ( ADB ).
The exterior angle ( angle ACD ) is equal to the sum of the two non-adjacent interior angles of triangle ( ADB ):
[tex]( \angle ACD = \angle DAB + \angle DBA ),[/tex]
[tex]( \angle ACD = 10^\circ + 20^\circ = 30^\circ ).[/tex]
[tex]∴ \text {The degree of} ( \angle ACD ) \: is \: ( 30^\circ ).[/tex]