Question

求大神解一下

求大神解一下第三大题三角函数

Answer 1

三、a=√2，c=2，A=30°，求b
解：sinC/c=sinA/a
sinC=csinA/a
=2sin30°/√2
=√2/2
C=45°、或者C=135°
C=45°时
B=180°-(30°+45°)
=105°
b/sinB=a/sinA
b=asinB/sinA
=√2sin105°/sin30°
=√2sin(60°+45°)/(1/2)
=2√2(sin60°cos45°+cos60°sin45°)
=2√2[(√3/2)(√2/2)+(1/2)(√2/2)]
=(2√2)(√2/2)(√3/2+1/2)
=√3+1
C=135°时
B=180°-(30°+135°)
=15°
b/sinB=a/sinA
b=asinB/sinA
=√2sin15°/sin30°
=√2sin(45°-30°)/(1/2)
=2√2(sin45°cos30°-cos45°sin30°)
=2√2[(√2/2)(√3/2)-(√2/2)(1/2)]
=(2√2)(√2/2)(√3/2-1/2)
=√3-1

Answer 2

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