Question

the side QR of a triangle pqr is produced to a point S if the bisector of angle pqr and angle PRS meet at point t then prove that angle QPR is equals to half angle QPR​

Answer 1

Step-by-step explanation

∠PRS is the exterior angle of ΔPQR

Therefore

∠PRS = ∠QPR + ∠PQR

⇒1/2∠PRS = 1/2∠QPR + 1/2∠PQR

⇒∠TRS = 1/2∠QPR + ∠TQR   ...............(1) {∵∠TRS =  1/2 ∠PRS and ∠TQR =

                                                                            1/2∠PQR

∠TRS is the exterior angle of ΔTQR

Therefore

∠TRS =∠QTR + ∠TQR    .........(2)

From equations (1) and (2)

∠QTR +∠TQR =1/2∠QPR + ∠TQR

⇒∠QTR =1/2 ∠QPR

Answer 2

$\underline\mathfrak{Given:-}$

• $\: \: \: \: \: \: \: \angle PRS \: \: = \: \ {132} \: \degree$
• $\: \: \: \: \: \: \: \angle RQP \: \: = \: \ {37} \: \degree$

$\underline\mathfrak{To \: \: Find:-}$

• $\: \: \: \: \: \: \angle QPR$
• $\: \: \: \: \: \: \angle PRQ$

$\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: \: \: \angle PRS \: + \: \angle PRQ \: \: = \: \: {180} \: \degree$

$\: \: \: \: \: \: \: \therefore \: Angle \: \: line \: \: property.$

$\: \: \: \: \: \: \: \leadsto {132} \degree \: + \: \angle PRQ \: \: = \: \: {180} \: \degree$

$\: \: \: \: \: \: \: \leadsto \angle PRQ \: \: = \: \: {180} \: \degree \: - \: {132} \degree$

$\: \: \: \: \: \: \: \leadsto \angle PRQ \: \: = \: \: {48} \: \degree$

$\: \: \: \: \: \: \: Now$

$\: \: \: \: \: \: \: In \: \: \triangle PQR$

$\: \: \: \: \: \: \: \leadsto \angle QPR \: + \: \angle PRQ + \angle PRQ \: \: = \: \ {180} \: \degree$

$\: \: \: \: \: \: \: \leadsto \angle QPR \: + \: {37} \: \degree \: + \: {48} \: \degree \: \: = \: \ {180} \: \degree \: \: \: \: \: \: {( Sum \: \: of \: \: angle \: \: of \: \: triangle.)}$

$\: \: \: \: \: \: \: \leadsto \angle QPR \: + \: {85} \: \degree \: \: = \: \ {180} \: \degree$

$\: \: \: \: \: \: \: \degree \angle QPR \: \: = \: \: {180} \: \degree \: - \ {85} \: \degree$

$\: \: \: \: \: \: \: \leadsto \angle QPR \: \: = \: \: {95} \: \degree \:$

$\: \: \: \: \: \: \: Hence,$

$\: \: \: \: \: \: \angle QPR \: \: \leadsto\: \: {95} \degree$

$\: \: \: \: \: \: \angle PRQ \: \: \leadsto \: \: {48} \degree$

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