Question

In the given figure, points P, Q, and R lie on the circumference of the circle centered at O. If $\angle OPQ$ measures $x^o$ and $\angle QRO$ measures $y^o$, calculate the measure of $\angle POR$ in terms of $x$ and $y$.

• A. $(360+x+y{)}^o$
• B. $(360-x-y{)}^o$
• C. $(360-2x-2y{)}^o$
• D. $(180+x+y{)}^o$

Answer 1

The correct option is C $(360-2x-2y{)}^o$

To solve the problem, we join points Q and O.

Now, in $\Delta OQP,OP=OQ=$ radius.

$\Rightarrow \angle OPQ=\angle OQP$

Since the sum of all angles in a triangle is ${180}^o,\angle POQ=(180-2x{)}^o$

Similarly, in $\Delta OQR,\angle OQR=\angle ORQ=y^o$

$\therefore \angle QOR=(180-2y{)}^o$

$\angle POR=\angle POQ+\angle QOR$

$\therefore \angle POR=(180-2x{)}^o+(180-2y{)}^o$

$=(360-2x-2y{)}^o$