Answer:
Question 1) [tex]x=6\°[/tex]
Question 2) [tex]x=9\°[/tex]
Question 4) [tex]arc\ BD=12\°[/tex]
Step-by-step explanation:
Question 1) we know that
The measure of the interior angle is the semi-sum of the arcs comprising it and its opposite
so
In this problem
[tex]124\°=\frac{1}{2}(arc\ RQ+arc\ SC)[/tex]
substitute the values and solve for x
[tex]124\°=\frac{1}{2}(197\°+(9x-3)\°)[/tex]
[tex]248\°=194\°+9x[/tex]
[tex]9x=248\°-194\°[/tex]
[tex]9x=54\°[/tex]
[tex]x=6\°[/tex]
Question 2) we know that
The measurement of the external angle is the semi-difference of the arcs it comprises
so
in this problem
[tex](5x-5)\°=\frac{1}{2}(190\°-(13x-7))[/tex]
solve for x
[tex](10x-10)\°=(197\°-13x)[/tex]
[tex](13x+10x)\°=(197\°+10\°)[/tex]
[tex]23x\°=207\°[/tex]
[tex]x=9\°[/tex]
Question 4) we know that
The measurement of the external angle is the semi-difference of the arcs it comprises
so
in this problem
[tex](6x-12)\°=\frac{1}{2}((-12+21x)\°-(12x-24))[/tex]
solve for x
[tex](12x-24)\°=((-12+21x)\°-(12x-24))[/tex]
[tex](24x-48)\°=(-12+21x)\°[/tex]
[tex](24x-21x)\°=(-12+48)\°[/tex]
[tex]3x=36\°[/tex]
[tex]x=12\°[/tex]
Find the measure of arc BD
[tex]arc\ BD=12x-24=12(3)-24=12\°[/tex]
