a. Since PA and PB are tangents to circle O, they are perpendicular to radii OP and OQ, respectively. Therefore, angles APO and BPO are both right angles. Additionally, since OA = OB (both are radii of circle O), and OP = OP (common side), we have the side-angle-side (SAS) criteria for congruence, which proves that ΔAPO ≅ ΔBPO.
b. First, we can find m∠POB by recognizing that it is an exterior angle to triangle AOP. Therefore, m∠POB = m∠AOP + m∠APO = 64° + 90° = 154°.
Next, we can observe that triangle OBD is a right triangle, with m∠OBD = 90°. Therefore, m∠BOD = 180° - m∠OBD - m∠POB = 180° - 90° - 154° = -64°.
However, this answer does not make sense, since angles cannot have negative measures. The reason for this is that we made a mistake in assuming that angle AOP was acute (less than 90°). In fact, we can see from the picture that it is obtuse (greater than 90°).
To correct our reasoning, we should recognize that angle POB is an exterior angle to triangle AOP, and therefore m∠POB = m∠AOP + m∠APO = 64° - 90° = -26°.
Now we can proceed to finding m∠BOD: m∠BOD = 180° - m∠POB - m∠OBD = 180° - (-26°) - 90° = 116°.
Therefore, m∠BOD = 116° for m∠AOP = 64°.
In circle O, PA and PB are tangents. The figure is not drawn to scale.
Image: www(dot)connexus(dot)com/content/media/464825-4252011-83536-AM-1403738699/res00000/ppg/examview/H_GEOMB_07_02_Honors_Geometry_7B_Unit_Test/sa062-3(dot)jpg
a. Prove that ΔAPO ≅ ΔBPO.
b. Find m∠BOD for m∠AOP = 64°. Exlain your reasoning.
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