
In the given figure, ST is a straight line and ray QP stands on it.

∴ \(∠\)PQS + \(∠\)PQR = 180º (Linear Pair)
\(∠\)PQR = 180º − \(∠\)PQS............ (1)
\(∠\)PRT + \(∠\)PRQ = 180º (Linear Pair)
\(∠\)PRQ = 180º − \(∠\)PRT............ (2)
It is given that \(∠\)PQR =\(∠\) PRQ.
Equating equations (1) and (2), we obtain
180º − \(∠\)PQS = 180º− \(∠\)PRT
\(∠\)PQS = \(∠\)PRT



Section | Number of girls per thousand boys |
|---|---|
Scheduled Caste (SC) | 940 |
Scheduled Tribe (ST) | 970 |
Non-SC/ST | 920 |
Backward districts | 950 |
Non-backward districts | 920 |
Rural | 930 |
Urban | 910 |
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
