In the figure, PS is the bisector of ∠QPR and PT ^ QR. Show that ∠TPS =
(∠Q - ∠R)
Since PS in the bisector of ∠QPR
... ∠QPS = ∠SPR
... ∠QPS =
∠P
Also in the right triangle PTQ, we have
∠Q +∠TPQ = 90° (... ∠PTQ = 90°)
... ∠TPQ = 90° - ∠Q
Now ∠TPS = ∠QPS - ∠QPT
= ∠QPR - (90° - ∠Q)
= ∠QPR - 90° + ∠Q
= [180° - (∠Q + ∠R)] - 90° + ∠Q
= 90° - ∠Q - ∠R - 90° + ∠Q = (∠Q - ∠R)
... ∠QPS = ∠SPR
... ∠QPS =
∠PAlso in the right triangle PTQ, we have
∠Q +∠TPQ = 90° (... ∠PTQ = 90°)
... ∠TPQ = 90° - ∠Q
Now ∠TPS = ∠QPS - ∠QPT
=
∠QPR - (90° - ∠Q)=
∠QPR - 90° + ∠Q=
[180° - (∠Q + ∠R)] - 90° + ∠Q= 90° -
∠Q - ∠R - 90° + ∠Q = (∠Q - ∠R) AI is thinking…
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.
∠A.
