Question

O is the centre of the circle. AB is a minor arc of the circle. The angle subtended by AB at centre, $\angle$AOB = ${110}^{\circ }$, then angle subtended by the arc at any point on the circle say $\angle$APB is ____, where P is any point on the circle.

• A. ${50}^{\circ }$
• B. ${45}^{\circ }$
• C. ${55}^{\circ }$
• D. ${65}^{\circ }$

Answer 1

The correct option is C ${55}^{\circ }$

As P is any point on the circle we can take that as shown in the figure.

Now we can easily see that in $\Delta$POB, OP=OB since both are radii of the circle.

So, $\angle$OPB = $\angle$OBP

As external angle is equal to sum of interior opposite angles

We can write $\angle$COB = $\angle$OPB + $\angle$OBP

= 2 $\angle$OPB

So, $\angle$COB = 2 $\angle$CPB

Similarly, in $\Delta$POA, OP=OA since both are radii of the circle.

So, $\angle$OPA = $\angle$OAP

As the external angle is equal to the sum of interior opposite angles.

We can write $\angle$COA = $\angle$OPA + $\angle$OAP

= 2 $\angle$OPA

So $\angle$COA = 2 $\angle$CPA

Hence, $\angle$COA+$\angle$COB = 2($\angle$CPA + $\angle$CPB)
Or, $\angle$ AOB = 2$\angle$APB
Therefore $\angle$APB = $\frac{1}{2}\angle AOB$
= $\frac{1}{2}\times {110}^o$
$\angle$APB = ${55}^{\circ }$.