ho givenOv2

1.	ho givenOv2
Answer» Step-by-step explanation: ∠PQR=90° (Angle in the semi circle is right angle) Also,from triangle PQR, we have ∠RPQ+∠PQR+∠PRQ=180° (angle sum property) ∠RPQ+90°+∠PRQ=180° ∠RPQ=90°-∠PRQ (1) Also, ∠BPR=90°(angle made on the tangent) (2) Now, ∠BPR=∠RPQ+∠QPB 90°=∠RPQ+∠QPB From (1), we get 90°=90°-∠PRQ+∠QPB ∠QPB=∠PRQ Hence proved

Answer»

Step-by-step explanation:

∠PQR=90° (Angle in the semi circle is right angle)

Also,from triangle PQR, we have

∠RPQ+∠PQR+∠PRQ=180° (angle sum property)

∠RPQ+90°+∠PRQ=180°

∠RPQ=90°-∠PRQ (1)

Also, ∠BPR=90°(angle made on the tangent) (2)

Now, ∠BPR=∠RPQ+∠QPB

90°=∠RPQ+∠QPB

From (1), we get

90°=90°-∠PRQ+∠QPB

∠QPB=∠PRQ

Hence proved

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