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ICSE Class X Math 2014 Paper

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In the figure, ∠DBC=58∘.BD\angle D B C = 58^{\circ} . B D is a diameter of the circle. Calculate:
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Question : 14 of 52
Marks: +1, -0
∠BEC\angle BEC
Solution:  
∠BEC+∠BDC=180∘\angle BEC + \angle BDC = 180^{\circ}
( ∵\because BECD is a cyclic quadrilateral)
∠BEC  =180∘−∠BDC\angle BEC \; = 180^{\circ} - \angle BDC
  =180∘−32∘\; = 180^{\circ} - 32^{\circ}
∠BEC  =148∘\angle BEC \; = 148^{\circ}
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