in the given figure pq is a tangent to the circle at a AND BD IS THE diameter.if∠abd=36° and ∠bdc=50°, calculate
∠qad,∠pab,∠cbd
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Answered by
23
Answer:
/_BCD = /_BAD = 90° [Angle in semicircle]
/_CBD + /_BDC + 90° = 180° [Angle Sum Property]
/_CBD = 40°
/_ADB + /_ABD + 90° = 180° [Angle Sum Property]
/_ADB = 54°
/_PAB = /_ADB [Alternate segment theorem]
/_PAB = 54°
/_QAD = /_ABD [Alternate segment theorem]
/_QAD = 36°
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Answered by
5
Answer:
angle QAD = 36
angle PAB =54
angle CBD = 40
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