Question

In the given figure, PT touches the circle whose center is O at R. Diameter SQ when produced meets PT at P. If angleSPR =x° and angleQPR=y°, show that x+2y=90°

Answer 1

Your ans is here ....

.

Answer 2

x+2y=90 is the Desired Operation

Step-by-step explanation:

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 $\times$ 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 is the desired operation !