Question

In the figure, $PT$ touches the circle at $R$ whose centre is $0$. Diameter $SQ$ when produced meets $PT$ at $P$. Given $\angle SPR=x^o$ and $\angle ORP=y^o$. Then,

• A. $x^o+2y^o={90}^o$
• B. $2x^o+y^o={90}^o$
• C. $x^o+y^o={120}^o$
• D. $3x^o+2y^o={120}^o$

Answer 1

The correct option is A $x^o+2y^o={90}^o$

$O$ is the centre of the circle.

$\angle SPR=x°\angle QRP=y°$

We know that $\angle QSR=\angle PRQ$ (angles in the alternate segment)

$=y°$

Also, $\angle QSR=90°$ ( a triangle with the base as the diameter of a circle and circumscribing it is right angled)

In $△QRS,$

$\angle SQR=180°-(\angle QRS+\angle QSR)=180°-90°-y°=90°-y°$

Now, $\angle SQR$ is an exterior angle of $△PQR$

$\angle SQR=\angle QPR+\angle PRQ=x°+y°=>90°-y°=x°+y°=>x+2y=90°$