Question

高一数学必修五 求解

Answer 1

一,
(1)cos∠ADC=1/7 ,sin∠ADC=√1-(cos∠ADC)^2]=4√3/7
sin∠ADB=sin∠ADC=4√3/7
cos∠ADB=-cos∠ADC=-1/7
sin∠BAD=sin(180-∠ABD-∠ADB)=sin(2π/3-∠ADB)=√3/2cos∠ADB+1/2sin∠ADB
=√3/2*(-1/7)+1/2*4√3/7
=3√3/14
所以sin∠BAD=11/14
(2)AB/sin∠ADB=BD/sin∠BAD
BD=(AB*sin∠BAD)/sin∠ADB=(8*3√3/14)/(4√3/7)=3
BD=3
(3)BC=BD+CD=3+2=5
AC^2=AB^2+BC^2-2AB*BCcosπ/3=8^2+5^2-2*8*5=9
AC=3
即有: sin∠BAD=11/14 , BD=5 ,AC=3
二
(1):(b-2a)cosC+ccosB=0
由正弦定理得:
(sinB-2sinA)cosC+sinCcosB=0
sinCcosB+sinBcosC-2sinAcosC=0
sin(C+B)-2sinAcosC=0
sinA-2sinAcosC=0
sinA不为0
1-2cosC=0
cosC=1/2
C=π/3
(2)c^2=a^2+b^2-2abcosπ/3
a^2+b^2-ab=c^2
a^2+(3a)^2-3a*a=(√7)^2
a=1 或 a=-1(舍去)
a=1 ,b=3a=3
S=1/2absinπ/3=1/2*1*3*√3/2=3√/4
S=3√/4
三
2asinA=(2b+c)sinB+(2c+b)sinC
由正弦定理得:
2a^2=b(2b+c)+c(2c+b)
2a^2=2b^2+bc+2c^2+bc
a^2=b^2+c^2+bc
b^2+c^2-a^2=-bc
cosA=(b^2+c^2-a^2)/2bc=-bc/2bc=-1/2
cosA=-1/2
A=2π/3