Question

I figure 5, sides QP and QR of ∆PQR are produced to the point S and T respectively if ∆SPR= 135° and ∆PQT= 110°, then find ∆PQR.​

Answer 1

Answer:65°

Step-by-step explanation:

Since, RQT is a straight line, so we can write that ∠PQT+∠PQR =180° ...... (1)

Given that ∠PQT=110°, hence, from equation (1), ∠PQR= 180°-110°=70°....... (2)

Since, QPS is a straight line, so we can write that ∠SPR+∠QPR =180° ...... (3)

Given that ∠SPR=135°, hence, from equation (3), ∠QPR= 180°-135°=45° .......(4)

Again in ΔPQR, ∠PQR +∠QPR +∠PRQ=180°

⇒70° +45° + ∠PRQ=180° {From equations (2) and (4)}

⇒ ∠PRQ =180°-70°-45° =65° (Answer)

Step-by-step explanation:

Answer 2

Answer:

65

Step-by-step explanation:

Since, RQT is a straight line, so we can write that ∠PQT+∠PQR =180° ...... (1)

Given that ∠PQT=110°, hence, from equation (1), ∠PQR= 180°-110°=70°....... (2)

Since, QPS is a straight line, so we can write that ∠SPR+∠QPR =180° ...... (3)

Given that ∠SPR=135°, hence, from equation (3), ∠QPR= 180°-135°=45° .......(4)

Again in ΔPQR, ∠PQR +∠QPR +∠PRQ=180°

⇒70° +45° + ∠PRQ=180° {From equations (2) and (4)}

⇒ ∠PRQ =180°-70°-45° =65° (Answer)

Step-by-step explanation: