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第1个回答 2018-08-20
A+3C=(π-B-C)+3C=π→2C=B
sinB/sinC=sin(2C)/sinC=2sinCcosC/sinC=2/√3→cosC=⅓√3
cosC=⅓√3→sinC=√(1-⅓)=√⅔→sinB=2sinCcosC=2√⅔·⅓=⅔√2
b/c=2/√3→c=4.5
S=½bcsinA=½·3√3·4.5sin(π-B-C)=6.75√3sin(B+C)
sin(B+C)=sinBcosC+sinCcosB=⅔√2·⅓√3+√⅔·√(1-8/9)
=⅓√6
∴S=6.75√3·⅓√6=6.75√2=
第2个回答 2018-08-20
(1)
A+3C=A+B+C
B=2C
由正弦定理得b/sinB=c/sinC
sinC=csinB/b=2sinCcosC/(b/c)=2sinCcosC/(2√3/3)
cosC=√3/3
(2)
sinC=√(1-cos²C)=√[1-(√3/3)²=√6/3
由正弦定理得
sinB=bsinC/c=(b/c)sinC=(2√3/3)(√6/3)=2√2/3
(3)
cosB=cos2C=1-2sin²C=1-2(√6/3)²=-⅓
sinA=sin(B+C)
=sinBcosC+cosBsinC
=(2√2/3)(√3/3)+(-⅓)(√6/3)
=√6/9
b=3√3
c=b/(2√3/3)=(3√3)/(2√3/3)=9/2
S△ABC=½bcsinA
=½·3√3·(9/2)·(√6/9)
=9√2/4
A+3C=A+B+C
B=2C
由正弦定理得b/sinB=c/sinC
sinC=csinB/b=2sinCcosC/(b/c)=2sinCcosC/(2√3/3)
cosC=√3/3
(2)
sinC=√(1-cos²C)=√[1-(√3/3)²=√6/3
由正弦定理得
sinB=bsinC/c=(b/c)sinC=(2√3/3)(√6/3)=2√2/3
(3)
cosB=cos2C=1-2sin²C=1-2(√6/3)²=-⅓
sinA=sin(B+C)
=sinBcosC+cosBsinC
=(2√2/3)(√3/3)+(-⅓)(√6/3)
=√6/9
b=3√3
c=b/(2√3/3)=(3√3)/(2√3/3)=9/2
S△ABC=½bcsinA
=½·3√3·(9/2)·(√6/9)
=9√2/4
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