Question

plz..solve my question
it is of 9th standard

Answer 1

<TRS + <TRP + <PRQ = 180 ( As RQ is produced so it beomes straight line

b+b + <PRQ = 180

<PRQ = 180 - 2b

As <Q = 2a

so <P + <Q + <PRQ = 180 ( sum of angles of triangle PQR is 180)

<P + 2a + 180 -2b = 180

<P = 2b - 2a

<P = 2( b-a)

and <P is nothing but <QPR

SO <QPR = 2( b-a)

In Triangle QTR

<TQR + <QTR +<TRQ = 180

TQR = a

TRQ = TRP + PRQ = b + 180 -2b = -b +180

So a + QTR -b +180= 180

QTR = b -a

So 2 <QTR = <QPR ( As <QPR = 2( b-a) as proved above)

<QTR = <QPR/2