question_answer

Two chords AB and CD of a circle with centre O, intersect each other at P. If \[\angle AOD=100{}^\circ \]and \[\angle BOC=70{}^\circ ,\] then the value of \[\angle APC\] is

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 \[\Delta APC\] \[\therefore \]      \[\angle APC=180{}^\circ -\angle CAB-\angle ACP\] \[=180{}^\circ -50{}^\circ -35{}^\circ =95{}^\circ \]