| In the following 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] |
| \[\therefore \] \[\angle OED=\angle ODE=\frac{180{}^\circ -\angle EOD}{2}\] |
| \[=\frac{180{}^\circ -80{}^\circ }{2}=50{}^\circ \]\[[\because OE=OD]\] |
| \[\therefore \] \[\angle ODC=180{}^\circ -\angle ODE\] |
| \[=180{}^\circ -50{}^\circ =130{}^\circ \] |
| Now, in \[\Delta DOC\] |
| \[\angle ODC+\angle DOC+\angle DCO=180{}^\circ \] |
| \[\Rightarrow \]\[130{}^\circ +15{}^\circ +\angle DCO=180{}^\circ \]\[\Rightarrow \]\[\angle DOC=35{}^\circ \] |
| \[\therefore \] \[\angle ECA=35{}^\circ \] |
| Alternate Method |
| Given, \[\angle BOD=15{}^\circ ,\]\[\angle EOA=85{}^\circ \] |
| \[\therefore \]\[\angle EOD=180{}^\circ -(85{}^\circ +15{}^\circ )=80{}^\circ \] |
| \[\therefore \]In \[\Delta EOD,\]\[EO=OD\] [radius of circle] |
| \[\therefore \]\[\angle OED=\angle ODE=\frac{(180{}^\circ -80{}^\circ )}{2}=\frac{100{}^\circ }{2}=50{}^\circ \] |
| Now, in\[\Delta OEC\] |
| \[\angle OCE=180{}^\circ -(50{}^\circ +80{}^\circ +15{}^\circ )\] |
| \[=180{}^\circ -145{}^\circ =35{}^\circ \] |
| \[\therefore \] \[\angle ECA=35{}^\circ \] |
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