Question

In fig 9.8 if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and $\angle BQR={70}^0$, then $\angle AQB$ is equal to

• A. ${20}^0$
• B. ${40}^0$,
• C. ${35}^0$,
• D. ${45}^0$

Answer 1

The correct option is B ${40}^0$,

$PQR$ is a tangent [ Given ]

$\therefore$ $\angle OQR={90}^o$ [ Angle between tangent and radius ]

$\Rightarrow$ $\angle OQB={90}^o-\angle BQR$

$\therefore$ $\angle OQB={20}^o$

Let the line from $O$ meet $AB$ at $M.$

Now In $△MQA$ and $△MQB$

$\Rightarrow$ $\angle QMA=\angle QMB$ [ Each ${90}^o$ ]

$\Rightarrow$ $\angle QM=QM$ [ Common side ]

$\Rightarrow$ $\angle MA=MB$ [ Perpendicular from center bisects the chord ]

$\therefore$ $△QMA\cong \angle QMB$ [ By $SAS$ rule ]

$\therefore$ $\angle MQA=\angle MQB={20}^o$

$\Rightarrow$ $\angle MQA+\angle MQB={20}^o+{20}^o$

$\Rightarrow$ $\angle AQB={40}^o$