| In the adjoining figure, AB is a diameter of the circle with centre O. If \[\angle BOD=15{}^\circ \]and \[\angle EOA=85{}^\circ ,\]then the measure of \[\angle ECA\]is [SSC (CPO) 2013] |
 |
A) \[45{}^\circ \]
B) \[35{}^\circ \]
C) \[30{}^\circ \]
D) \[70{}^\circ \]
Correct Answer: B
Solution :
| \[\angle AOB=180{}^\circ \] |
| \[\Rightarrow \]\[\angle AOE+\angle EOD+\angle DOB=180{}^\circ \] |
| (by angle sum property) |
| \[\Rightarrow \]\[85{}^\circ +\angle EOD+15{}^\circ =180{}^\circ \] |
| \[\angle EOD=80{}^\circ \] |
| \[\because \] \[OD=OE\] [radius of circle] |
| \[\therefore \]\[\angle OED=\angle ODE=\frac{180{}^\circ -\angle EOD}{2}\] |
| \[=\frac{180{}^\circ -80{}^\circ }{2}=50{}^\circ \] \[[\because OE=OD]\] |
| \[\therefore \]\[\angle ODC=180{}^\circ -\angle ODE=130{}^\circ \] |
| Now, in \[\Delta DOC,\]\[\angle ODC+\angle DOC+\angle DCO=180{}^\circ \] |
| \[130{}^\circ +15{}^\circ +\angle DCO=180{}^\circ \] |
| \[\Rightarrow \] \[\angle DCO=35{}^\circ \] |
| \[\angle ECA=35{}^\circ \] |
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