Question

在三角形ABC中,∠ACB是直角,∠B=60,AD,CE分别是∠BAC和∠BCA的平分线,AD,CE相交于点F,求∠EF

Answer 1

∵∠ACB=90° ∠B=60°

∴∠BAC=30°

∵平分线

∴∠CAF=1/2∠BAC=15°

∠ACF=1/2∠ACB=45°

∵∠EFA=∠CAF+∠ACF

∴∠EFA=60°