Question

已知a，β为锐角，cosa=1/7,sin(a+β)=5√3/14,求角β的值

Answer 1

解：
已知α，β为锐角，
所以sinα>0 cosβ>0 sinβ>0
cosα=1/7
sinα=√(1-cos²α)=√(1 - 1/7²)= (4√3)/7

sin(α+β)
= sinαcosβ+cosαsinβ
= (4√3)/7cosβ+(1/7)sinβ
= (4√3)/7√(1-sin²β) +(1/7)sinβ
即
(4√3)/7√(1-sin²β) +(1/7)sinβ = 5√3/14
(4√3)/7√(1-sin²β) = 5√3/14 -(1/7)sinβ
两边平方

(48/49)(1-sin²β)= 75/196 - (5√3)/49sinβ + (1/49)sin²β
sin²β- (5√3)/49sinβ- 117/196 = 0
(sinβ- (5√3)/98)² = 5808/9604
sinβ- (5√3)/98 = (44√3)/98
sinβ = (49√3)/98 = (√3)/2
β = 60°

Answer 2

貌似无解