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Consider circle O, where m∠BD = 70° and m∠CA = 170°. m∠BPD = ____°. A. 100° B. 110° C. 120° D. 130°

User Obaa
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Final answer:

m∠BPD is 120° because it is an inscribed angle intercepting an arc with a central angle (angle BOD) measure of 240° in circle O. As such, it is half the measure of angle BOD.

Step-by-step explanation:

We are asked to find the measure of angle BPD in circle O, given that m∠BD = 70° and m∠CA = 170°. To find this measure, we can apply the facts that the sum of the measures of the angles in a quadrilateral is 360° and that an inscribed angle in a circle is half the measure of its intercepted arc.

In circle O, angle CA and angle BD are inscribed angles that intercept arcs CB and AD, respectively. Since m∠CA = 170°, the measure of arc CB is twice that, which is 340°. Now, seeing that the total degrees in a circle add up to 360°, the remaining arc, AD, would have a measure of 360° - 340° = 20°. Because angle BD is an inscribed angle intercepting arc AD, m∠BD = 70° is half of its intercepted arc, confirming the intercepted arc AD is 20°, as calculated.

Now, angle BPD is an exterior angle at point B to the quadrilateral ABDC, therefore its measure is equal to the sum of the opposite interior angles, which are CA and BD (170° + 70°) = 240°. Now, considering that vertical angles are congruent and angle BPD and angle BOD are vertical angles, the measure of angle BOD (central angle) is also 240°. Thus, m∠BPD is half the measure of angle BOD because it is an inscribed angle that intercepts the same arc, which is 120°.

User Amit Thakur
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