O:250 degrees
A
In this figure, mzBDA =
• and m2BCA =
Please
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Answer:
∠ BDA = 55° , ∠ BCA = 70°
Step-by-step explanation:
∠ BOA = 360° - 250° = 110° ( angles round a point )
The central angle is twice the angle on the circle, subtended on the same arc
∠ BDA = [tex]\frac{1}{2}[/tex] ∠ BOA = [tex]\frac{1}{2}[/tex] × 110° = 55°
OBCA is a tangent kite
Opposite angles sum to 180° , then
∠ BCA = 180° - ∠ BOA = 180° - 110° = 70°