In the given figure, PA is a tangent from an external point P to a circle with centre O. If $\angle\text{POB} = 125^\circ$, then $\angle\text{APO}$ is equal to :
Show Hint
Using the exterior angle theorem for $\Delta\text{OAP}$:
The exterior angle $\angle\text{POB}$ is equal to the sum of the two interior opposite angles:
\[ \angle\text{POB} = \angle\text{OAP} + \angle\text{APO} \]
Since $\angle\text{OAP} = 90^\circ$, we have:
\[ 125^\circ = 90^\circ + \angle\text{APO} \implies \angle\text{APO} = 125^\circ - 90^\circ = 35^\circ \]
This method is much faster as it avoids calculating the linear pair!
Step 1: Set up the right angle at the point of contact. Since PA is tangent to the circle at A, the radius OA is perpendicular to PA, so $\angle OAP=90^\circ$. Step 2: Use the exterior angle theorem on triangle OAP. $\angle POB$ is the exterior angle of $\Delta OAP$ at O, so it equals the sum of the two interior opposite angles: $\angle POB=\angle OAP+\angle APO$. Step 3: Solve for the unknown angle. $125^\circ = 90^\circ+\angle APO \implies \angle APO = 35^\circ$. \[ \boxed{35^\circ} \]
