Question

如图,若△ABC的外角∠DBC,∠BCE的平分线相交于点O,∠A=40°,求∠BOC的度数

Answer 1

∠BOC=180°-∠3-∠4

=180°-1/2∠DBC-1/2∠ECB

=180°-1/2(∠DBC+∠ECB)

=180°-1/2(180°-∠ABC+180°-∠ACB)

=180°-1/2*(180°+∠A)

=90°-1/2∠A

=90°-1/2x40°

=90°-20°

=70°