在△ABC中,A=45度.B=60度.AB=2,求△ABC的面积.(成考09.24)请用正弦定理.可知(BC/SINA)=(AB/SINC),则BC=(2X√2)/SIN(45+60)=(√2)/(√2/2X√3/2+√2/2+1/2)我不明白接着为什么等于(√2)/(√6+√2),下面还有分母2哪里

Question

在△ABC中,A=45度.B=60度.AB=2,求△ABC的面积.(成考09.24)
请用正弦定理.
可知(BC/SINA)=(AB/SINC),则
BC=(2X√2)/SIN(45+60)=(√2)/(√2/2X√3/2+√2/2+1/2)
我不明白接着为什么等于 (√2)/(√6+√2),下面还有分母2哪里去了.
=2(√3-1)

Answer 1

sin(a+b)=sinacosb+cosasinb
所以sin(45+60)
=sin45cos60+cos45sin60
=√2/2×√3/2+√2/2×1/2
=(√6+√2)/4
sinA=√2/2
所以这里分子漏了1/2
即BC=AB*sinA/[(√6+√2)/4]
=2×√2/2/[(√6+√2)/4]
=4√2/(√6+√2)
=4/(√3+1)
=2(√3-1)