Question

In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.

Given $\angle SPR=x^o$ and $\angle QRP=y^o$

prove that:

(i) $\angle ORS=y^o$

(ii) write an expression connecting x and y.

Answer 1

given-
QRP=y
ORP=90 (radius to tangent through pt of contact)
ORP-QRP=90-y=QRO

now,
QRS=90 (ANGLE IN SEMICIRCLE)
QRO+ORS=QRS
​(90-y)+ORS=90
ORS=90-(90-y)
ORS=y
HENCE PROVED
ii)
ORS=OSR=y (ANGLES OPP.TO EQUAL SIDES-RADII)
QOR=2*OSR (ANGLE SUBTENTED AT THE CENTRE BY A CHORD IS TWICE THE ANGLE SUBTENTED BY IT ON ANY OTHER PT ON THE CIRCLE)
QOR=2y
OQR=QPR+QRP (EXT. ANGLE PROP)
OQR=x+y
OQR=ORQ=x+y (ANGLES OPP. TO EQUAL SIDES)
OQR+ORQ+QOR=180 (ANGLE SUM PROP)
x+y+x+y+2y=180
2x+4y=180
x+2y=90