Question

angle q is greater than angle r .ps is bisector of angle pqr . pm is perpendicular to QR. find angle mps

Answer 1

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Given: In ΔPQR, ∠ Q > ∠ R. If PA is the
bisector of ∠ QPR and PM perp QR.
To Prove: ∠APM =(1/2) ( ∠ Q - ∠ R )
Proof: Since PA is the bisector of ∠P,we have,
∠APQ=(1/2) ∠P....................(i)
In right -angled triangle PMQ,we have,
∠Q+ ∠MPQ=90°
⇒ ∠MPQ= 90°-∠Q...................(ii)
∴∠APM=∠APQ-∠MPQ
1/2 ∠P - (90 - ∠Q) [using (i) and (ii)]
1/2∠P-90+∠Q
1/2∠P - 1/2(∠P + ∠R + ∠Q ) +∠Q [since 90
= 1/2(∠P + ∠R + ∠Q)]
1/2(∠Q -∠R)

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