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                                                                                                                                                                                        [SSC (CGL) Mains 2014]

A) \[80{}^\circ \]                          

B) \[75{}^\circ \]

C) \[85{}^\circ \]                          

D) \[95{}^\circ \]

Correct Answer: D

Solution :

(d) In the given figure,

\[\angle AOD=100{}^\circ \]\[\Rightarrow \]\[\angle BOC=70{}^\circ \]

Now, join AC.

\[\angle ACD=\frac{1}{2}\angle AOD\]

[since, angle subtended at the centre is twice the angle subtended on circumference of following circle]

\[=\frac{1}{2}\times 100=50{}^\circ \]

Similarly, \[\angle CAB=\frac{1}{2}\times \angle DOB=\frac{1}{2}\times 70{}^\circ =35{}^\circ \]

In \[\Delta APC,\]\[\angle APC=180{}^\circ -\angle ACP-\angle CAB\]

\[=180{}^\circ -50{}^\circ -35{}^\circ \]\[[\because \angle ACP=\angle ACD]\]\[=95{}^\circ \]