若△ABC是钝角三角形,求arccos(sinA)＋arccos(sinB)＋arccos(sinC)的取值范围.（90°,270°）2

Question

若△ABC是钝角三角形,求arccos(sinA)＋arccos(sinB)＋arccos(sinC)的取值范围.（90°,270°） 2

Answer 1

设：A为钝角
∵B,C都小于90°
∴arccos(sinA)=A-90°
arccos(sinB)=90°-B
arccos(sinC)=90°-C
arccos(sinA)＋arccos(sinB)＋arccos(sinC)
=(A-90°)＋(90°-B)＋(90°-C)
=90°+A-B-C
∵A-B-C的值的范围是0°到180°
∴arccos(sinA)＋arccos(sinB)＋arccos(sinC)的取值范围是：（90°,270°）