三角函數題目

從題2判斷題2&3應該是分母分子寫顛倒了

1.｜a+2b-2c｜+｜3a-2c｜=0
=>a+2b-2c=0----(1)
3a-2c=0 ----------(2)
由(2) 得 a =2/3 c代入(1)
(2/3 c)+2b -2c=0 => b= 2/3 c
=> a: b: c =2/3 c : 2/3 c : c =2 : 2: 3
sinA:sinB:sinC= a: b: c = 2 : 2: 3

2.△ABC中，AD⊥BC，已知AB=15,sinB=3/5 cotC=1/3, 則BC=?
Sol: sinB=BD/AB=3/5 =>AD/15=3/5 => AD=9
=> BD =√(AB^2-AD^2) = 12
cotC=1/3 = CD/AD => CD/9= 1/3 => CD=3
則BC=BD+CD= 12+3=15

3. 1+tanθ / 1- tanθ = 3+2√2
=>1+tanθ= (3+2√2)- (3+2√2)tanθ
=>(4+2√2)tanθ= (2+2√2)
=> tanθ = 1 / √2
=> sinθ = 1/√3 , cosθ= √2 /√3
sinθ+cosθ= √2+1 /√3 = (√6+√3)/3

2,3題，題目錯誤．．．．．．．