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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 : 13 of 52
Marks: +1, -0
∠BDC\angle BDC
Solution:  
In △BCD;∠DBC=58∘\triangle BCD ; \angle DBC=58^{\circ} (given)
∠BCD=90∘\angle BCD=90^{\circ}
(Angle in the semicircle as BDBD is diameter)
∴∠DBC+∠BCD+∠BDC=180∘\therefore \angle DBC+\angle BCD+\angle BDC=180^{\circ}
58∘+90∘+∠BDC=180∘58^{\circ}+90^{\circ}+\angle BDC=180^{\circ}
⇒∠BDC=180∘−(90∘+58∘)\Rightarrow \qquad \angle BDC=180^{\circ}-(90^{\circ}+58^{\circ})
=180∘−148∘=32∘=180^{\circ}-148^{\circ}=32^{\circ}
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