If 0

If 0<A<B<π, sinA-sinB=12cosA-cosB=32, then A+B=

  1. A

    2π3

  2. B

    5π6

  3. C

    π

  4. D

    4π3

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    Solution:

    (sinA-sinB)2+(cosA-cosB)2=12+32=2

    2(sinAsinB+cosAcosB)=0

    cos(B-A)=0B=A+π2

    Now, sinA - sinB =12sinA-cosA=12

     sinA2-cosA2=12 sinA-π4=sinπ6A-π4=π6A=5π12  A+ B =A+ A+π2=5π6+π2=4π3

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