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Liltico

  • 2 years ago

tan 49 deg = 19/x

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  1. Liltico
    • 2 years ago
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    @BTaylor

  2. Liltico
    • 2 years ago
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    i can't find the answer for x :(

  3. agentx5
    • 2 years ago
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    \(\large \tan \theta = \huge \frac{opposite}{adjacent} \)

  4. agentx5
    • 2 years ago
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    Algebra tricks... (writes)

  5. agentx5
    • 2 years ago
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    \[\tan 49^o = \frac{19}{x}\] \[x\tan 49^o = 19\] \[x = \frac{19}{\tan 49^o}\]

  6. Liltico
    • 2 years ago
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    16.51?

  7. Qibtiya
    • 2 years ago
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    @Liltico did u get it?

  8. Liltico
    • 2 years ago
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    16.52

  9. agentx5
    • 2 years ago
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    @Liltico is correct.

  10. Liltico
    • 2 years ago
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    thanks agent

  11. agentx5
    • 2 years ago
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    You got a typo I see @Qibtiya , it's 49\(^o\) not 14\(^o\)

  12. Qibtiya
    • 2 years ago
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    oh i thought that tan is 14 degree instide of 49 tan 49=1.150 tan 49=19/x or x=19/tan 49 put the value of tan 14 x=19/1.150 x=16.51

  13. Qibtiya
    • 2 years ago
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    @agentx5 yeah got it

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