• 2021 April(D05) Math Question 37

    Trigonometry

    - Pythagorean Theorem & sin cos tan

    Pythagorean Theorem and Special Right Triangles, Trigonometry for Right Triangles
    Choice D
    JL={92+72}={130}JL=\sqrt\{9^2+7^2\}=\sqrt\{130\}
    ∴cos  KJL={KJ}{JL}={9}{{130}}\therefore cos\; KJL=\frac\{KJ\}\{JL\}=\frac\{9\}\{\sqrt\{130\}\}

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