Question

在矩形ABCD中,AC、BD交于点O,AE平分∠BAD,若∠EAO=15°,求∠AOE的度数

Answer 1

因为AE平分∠BAD

所以∠BAE = 1/2∠BAD = 45°

∠BEA = 90°- ∠BAE = 90°- 45°= 45°

∠DAC = ∠DAE - ∠EAO = 45°-15°= 30°

设边长AB = 1

得到 AE = √2

AO = 1

解三角形AOE

余弦定理有 EO^2 = AO^2 + AE^2 - 2AO*AE*cos∠EAO

= 1 + 2 -2√2cos15°

= 3-2√2cos15°

EO = √(3-2√2cos15°)

正弦定理 EO/sin∠EAO = AE/sin∠AOE

sin∠AOE = AE*sin∠EAO/EO

= √2sin15°/ √(3-2√2cos15°)

所以 ∠AOE = arc sin( √2sin15°/ √(3-2√2cos15°) )