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.

1 answer

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°.