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anonymous

  • 4 years ago

Given that sin θ = sqrt(3)/7 and tan θ < 0, find sec θ and tan θ.

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  1. y2o2
    • 4 years ago
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    θ is in the second quad. so you can use sin^-1 function or you can use Pythagoras theorem and SOH CAH TOA

  2. anonymous
    • 4 years ago
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    ok so x^2 + y^2 = r^2 x^2 + sqrt(3)^2 = 7^2 x^2 + 3 = 49 x^2 = 46 x = sqrt(46) right?

  3. y2o2
    • 4 years ago
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    yes you got the adjacent right. so now you can get the cos and the tan but remember that both of them will be -ve

  4. anonymous
    • 4 years ago
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    So Sec θ = - 7sqrt(46)/46 Tan θ = - sqrt(138)/46

  5. y2o2
    • 4 years ago
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    Yes, that's right

  6. anonymous
    • 4 years ago
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    cool, thanks. I wasn't sure if I could reduce it further.

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