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ccsoccerox

  • 3 years ago

Given the triangle below, what is sec ∡B? a. 4 sqrt33/33 b. 7/4 c. 4/7 d. 7 sqrt33/33

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  1. ccsoccerox
    • 3 years ago
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  2. WATTAGWAN
    • 3 years ago
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    Sec B = 1/ Cos B and cosB = A/7 so sec B = 7/A where A = \[\sqrt{7^2 - 4^2}\]

  3. WATTAGWAN
    • 3 years ago
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    Seriously? you can do the rest with a calculator.

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