Question

In the figure above AB is a diameter of the circle with the centre O, AP and AQ are equal chords, If $\angle PAQ={80}^{\circ }$ then find $\angle APO$

• A. ${45}^{\circ }$
• B. ${40}^{\circ }$
• C. ${80}^{\circ }$
• D. ${100}^{\circ }$

Answer 1

The correct option is B ${40}^{\circ }$

In $\Delta AOP,AO=OP$

$\therefore$ $\angle APO=\angle OAP⟶\left(1\right)$

$\angle PAQ={80}^0\Rightarrow \angle PAO+\angle OAQ={80}^0⟶\left(2\right)$

Now, in $\Delta PAO\&\Delta OAQ$

(i) $AP=AQ$ [equal chords] (ii) $OP=OQ$ [Radii]

(iii) $AO=AO$ [common] Therefore $\Delta PAO\cong \Delta QAO$ [By sss]

$\therefore$ $\angle OAP=\angle OAQ$ [By CPCT]

from (ii), $\angle OAP+\angle OAP={80}^0\Rightarrow 2\angle OAP={80}^0$

$\Rightarrow \angle OAP={40}^0$

Therefore, from (i), we get

$\angle APO={40}^0$