2)
2cos2B-8cosB+5=0
→? 2(2cosB^2- 1)-8cosB+5=0?
→? Let t=cosB
then what The original equation is transformed into? 4t^2-8t+3=0
→? t= 1/2? Or 3/2? CosB= 1/2。
→? B=60
a*b=|a|*|b|=3*5*cosQ=-9? →? cosQ=-3/5? sinQ=4/5
done
sin(B+Q)
=sinBcosQ+cosBsinQ
=√3/2*(-3/5)+ 1/2*4/5
=(4-3√3)/ 10
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