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 $\angle$APB is

[P is any point on the circle]

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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

Now, we can easily see that in $△$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 $△$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
Substituting, we can get $\angle$APB $={55}^{\circ }$.