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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