作业帮高一数学,正弦定理第9题
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高一数学,正弦定理第9题
高一数学,正弦定理第9题
高一数学,正弦定理第9题
解析：本题考查知识点：正弦定理,两角和与差的正弦由正弦定理a/sinA = b/sinB = 2R知：a = 2RsinA,b = 2RsinB
∵C=60°∴A+B = 120°,∴B=120°-A∴(a+b)/R = (2RsinA+2RsinB)/R
= 2R(sinA + sinB)/R= 2(sinA+sinB)=2[sinA+sin(120°-A)]=2(sinA+sin120°cosA-cos120sinA)=2(sinA+√3/2cosA+1/2sinA)=2(3/2sinA+√3/2cosA)=2√3(√3/2 sinA+1/2 cosA)=2√3sin(A+60°)∵0
二分之根3*R~根3*R，根号不会打，刚刚搞错了少写了个字，希望没误导你。
正弦定理：
a/sinA = b/sinB = 2R
a = 2RsinA
b = 2RsinB
(a+b)/R = (2RsinA+2RsinB)/R
= 2R(sinA + sinB)/R
= 2(sinA+sinB)
= 2(2 * sin[(A...
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正弦定理：
a/sinA = b/sinB = 2R
a = 2RsinA
b = 2RsinB
(a+b)/R = (2RsinA+2RsinB)/R
= 2R(sinA + sinB)/R
= 2(sinA+sinB)
= 2(2 * sin[(A+B)/2] * cos[(A-B)/2])
= 2(2 * sin[(180°-C)/2] * cos[(A-B)/2])
= 2(2 * sin60° * cos[(A-B)/2])
= 2√3cos[(A-B)/2]
∵A+B = 120°
∴A-B = 120° - 2B
2√3cos[(A-B)/2] = 2√3cos(60°-B) = 2√3cos(B-60°)
∵C=60°
∴0
1/2 < cos(B-60°) ≤ 1
√3 < 2√3cos(B-60°) ≤ 2√3
所以取值范围：
(√3 , 2√3]
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