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tan 49 deg = 19/x
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@BTaylor
i can't find the answer for x :(
\(\large \tan \theta = \huge \frac{opposite}{adjacent} \)
Algebra tricks... (writes)
\[\tan 49^o = \frac{19}{x}\] \[x\tan 49^o = 19\] \[x = \frac{19}{\tan 49^o}\]
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16.51?
@Liltico did u get it?
16.52
@Liltico is correct.
thanks agent
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You got a typo I see @Qibtiya , it's 49\(^o\) not 14\(^o\)
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
@agentx5 yeah got it
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