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Albert Singh Bandi
Asked: 2022-11-10T08:45:17+05:30 In:

If `sinalpha+cosbeta=2,(0^(@) le beta lt alpha90^(@))`, then `sin((2alpha+beta)/(3))=?`
A. `sin((alpha)/(2))`
B. `cos((alpha)/(3))`
C. `sin((alpha)/(3))`
D. `cos((2alpha)/(3))`

If `sinalpha+cosbeta=2,(0^(@) le beta lt alpha90^(@))`, then `sin((2alpha+beta)/(3))=?`
A. `sin((alpha)/(2))`
B. `cos((alpha)/(3))`
C. `sin((alpha)/(3))`
D. `cos((2alpha)/(3))`
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1 Answer

  1. Correct Answer – b
    `sinalpha+cos beta= 2`
    shortest method
    put, `alpha=90^(@),beta=0^(@)`
    `sin90^(@)+cos0^(@)=2`
    `1+1=2`
    2=2matched
    So, `alpha=90^(@),beta=0^(@)`
    `rArrsin((2alpha+beta)/(3))^(2)`
    `=sin((2xx90+0)/(3))^(2)`
    `=sin((180)/(3))^(2)`
    `=sin60^(@)=cos30^(@)=sqrt(3)/(2)`
    Take `cos(alpha)/(3)=cos(90^(@))/(2)`
    `=cos30^(@)`
    So, this is answer

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