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प्रश्न
In the joining figure shown XAY is a tangent. If ∠ BDA = 44°, ∠ BXA = 36°.
Calculate: (i) ∠ BAX, (ii) ∠ DAY, (iii) ∠ DAB, (iv) ∠ BCD.
उत्तर
(i) ∠ BAX = ∠ BDA = 44° ...(Angles in the alternate segment)
(ii) ∠ ABD = ∠ BXA + ∠ BAX
∠ ABD = 36° + 44° = 80° ....(Ext. angle of a triangle = sum of internal opposite angles)
∴ ∠ DAY = ∠ ABD = 80° ...(Angles in the alternate segment)
(iii) ∠ DAB = 180° - (∠ BAX + ∠ DAY)
∠ DAB = 180° - (44° + 80°) = 56°
(v) ∠ BCD = 180° - ∠ DAB
∠ BCD = 180° - 56° = 124° ...(Opposite ∠ s of a cyclic quadrilateral are supplementary)
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