| Let ABCD is a convex quadrilateral in which \[\angle BAC=50{}^\circ ,\] \[\angle CAD=60{}^\circ ,\] \[\angle CBD=30{}^\circ \] & \[\angle BDC=25{}^\circ .\] If E is the point of intersection of AC & BD then \[\angle AEB\] equals - |
 |
A) \[65{}^\circ \]
B) \[75{}^\circ \]
C) \[85{}^\circ \]
D) \[95{}^\circ \]
Correct Answer: D
Solution :
| \[\because \] A is center of circle (as shown in figure) because chord CD & BC subtends twice the angle at A w.r.t. point B & D respectively. |
| \[\because \,\,\,\angle ABE=\angle ADE=\frac{180{}^\circ -110{}^\circ }{2}=35{}^\circ \] |
| \[\therefore \,\,\,\angle AEB=180{}^\circ -(50{}^\circ +35{}^\circ )=95{}^\circ \] |
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