Question

一道数学题目求解，17题

Answer 1

2sin²(π/4＋B/2)=1－cos(π/2＋B)=1＋sinB
m⊥n，则：m*n=0，得：
2sinB(1＋sinB)＋(cos2B－2)=0
2sinB＋2sin²B＋2cos²B－3=0
2sinB=1
sinB=1/2
B=30°或B=150°

c²=a²＋b²－2abcosB
(1)若B=30°，则：c²=1，得：c=1
(2)若B=150°，则：c²=7，得：c=√7

Answer 2

解：（1）∵m⊥n ∴(cos2B-2)/(2sinB)(1+sinB) = -1
2sinB + 2sin²B = 1+2sin²B
2sinB = 1
sinB = 1/2
∴ B = π/6 或 B = 5π/6

（2）若 a = √3 ， b = 1
sinA = asinB/b = √3/2
又∵b < a
∴ B为锐角，此时 B = π/6
A = π/3 或 A = 2π/3
当 A = π/3 时 C = π/2，此时：c = 2
当 A = 2π/3时 C = π/6 ，此时：c = b = 1

Answer 3

（1）m⊥n得：2sinB(2sin²(π/4+B/2))-(2-cos2B)=0;
即：2sinB[1-cos(π/2+B)]-2+cos2B=0; 2sinB+2sin²B-2+1-2sin²B=0
则：sinB=1/2; B=30°,或B=150°
(2)
当a=√3>b=1时，B=30°;
则a/sinA=b/sinB得：sinA=√3/2; A=60°或A=120°
则C=180°-(A+B)=90°或30°;
从而c=2或c=1

Answer 4

m*n=0=2sinB*2sin²(π/4+B/2)-2+cos2B=2sinB*(1-cos(B+π/2))-2+cos2B
=2sinB*(1+sinB)-2+2cosB*cosB-1=2sinB-1
sinB=1/2 B=30或150
2.a/sina=b/sinb a=60或120 显然b不能150只能30度
b=30 a=60度是c=90 c=2
a=120 b=30 c=30 c=1