Question

已知直角三角形abc中,角c=90度,角a,角b的角平分线交于点p,求角apb的度数

Answer 1

135°

∵直角三角形ABC

∴∠C = 90°

∴∠BAC+∠CBA=90°

∵AD平分∠BAC

∴∠CAD=∠BAD=12∠CAB

∵BE平分∠CBA

∴∠CBE=∠ABE=12∠CBA

∴∠ DAB+∠EBA=12∠CAB+12∠CBA

=12(∠CAB+∠CBA)

=12×90 °

=45°

∴∠ APB=180°-∠DAB-∠EBA

=180°-(∠DAB+∠EBA)

=180°-45°

=135°