Question 11 Mark
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, $\angle$POA is equal to
Answer
View full question & answer→Here $\angle $APB = 80°
$\therefore $$\angle $AOB = 180° - 80° = 180°
Now, since OP bisect $\angle $APB and $\angle $AOB.
$\therefore $$\angle $AOP = $\frac{{{{100}^ \circ }}}{2} = {50^ \circ }$
$\therefore $$\angle $AOB = 180° - 80° = 180°
Now, since OP bisect $\angle $APB and $\angle $AOB.
$\therefore $$\angle $AOP = $\frac{{{{100}^ \circ }}}{2} = {50^ \circ }$

