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es12.In the given figure, if <BAD-60 and <ADC-105° then determine <DPC3X10-30 |
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Answer» Given--- ∠ ADC = 105° We know,∠ADC +∠ABC = 180° [Opposite angles of the Cyclic Quadrilateral are Supplementary] ∠ABC ( or ∠PBA) = 180° - 105° = 75°∠DAB( or∠PAB) = 60° [Given] Using the Angle Sum Property inΔPAB, ∠PAB + ∠PBA +∠APB = 180° 60° + 75° +∠APB = 180° ∠APB = 180° - 135° ∠APB = 45° Thus, ∠DPC(or ∠APB) is 45°. |
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