The diagram is missing but I'll assume that the arc BDC is:
[tex](4k + 159)^{\circ}[/tex]
And another arc, let's call it FGH. measures:
[tex](2k + 153)^{\circ}[/tex]
If those arc are equal, then this equation is true:
[tex](4k + 159)^{\circ}=(2k + 153)^{\circ} \\ \\ (4k + 159)=(2k + 153) \\ \\ \\ Solving \ for \ k: \\ \\ 4k-2k=153-159 \\ \\ 2k=-6 \\ \\ k=-\frac{6}{2} \\ \\ k=-3[/tex]
Substituting k into the first equation:
[tex]\angle BDC=(4(-3)+159)^{\circ} \\ \\ \angle BDC=(-12+159)^{\circ} \\ \\ \boxed{\angle BDC=147^{\circ}}[/tex]
Answer:
The answer is 213, because 360-147=213 and they are asking you for BDC a 360 degree angle
Step-by-step explanation:
