A) \[80{}^\circ \]
B) \[75{}^\circ \]
C) \[85{}^\circ \]
D) \[95{}^\circ \]
Correct Answer: D
Solution :
[d] In the given figure \[\angle AOD=100{}^\circ \] \[\angle COB=70{}^\circ \] Now, join AC \[\angle ACP=\frac{1}{2}\angle AOD\] [angle subtended at the centre is twice the angle subtended on circumference of following circle] \[=\frac{1}{2}\times 100{}^\circ =50{}^\circ \] Similarly, \[\angle CAB=\frac{1}{2}DOB=\frac{1}{2}70{}^\circ \] \[=35{}^\circ \] In \[\Delta APC\] \[\therefore \] \[\angle APC=180{}^\circ -\angle CAB-\angle ACP\] \[=180{}^\circ -50{}^\circ -35{}^\circ =95{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec