Question

AB是半圆O的直径,点C在半圆上,且CO^2=CA*CB,则角CAB=?

Answer 1

CA*CB=2S(ABC)

S(ABC)=1/2*AB*h=1/2*(2OC)*(OCsinθ)=OC^2*sinθ

2S=2OC^2sinθ

and 2S=CA*CB=CO^2

so

sinθ=1/2

=>θ=30 ,or 150

θ is ∠COA

when ∠COA=30 ,∠CAB=75

when ∠COA=150,∠CAB=15

so conclusion is 15