An observer stationed 20 m away from a tall

Mathematics

2 answers:

antiseptic1488 [7]3 years ago

5 0

Answer:

Step-by-step explanation:I don't say you must have to mark my ans as brainliest but my friend if it has really helped you plz don't forget to thank me...

Nimfa-mama [501]3 years ago

4 0

Alazgggggg akstafarrrallaa ?

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x=[-5(+or-)sqrt(5^2-4*1*5)]/(2*1)
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beks73 [17]

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Because I said so

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Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))

miv72 [106K]

$\bf tan\left[ 2cos^{-1}(x) \right]\implies tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]
\\\\\\
\textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{ that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\
-----------------------------\\\\
cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\
-----------------------------\\\\
$

$\bf \textit{again, using the pythagorean theorem to get the opposite side}
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c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b
\\\\\\
\pm\sqrt{1-x^2}=b\\\\
-----------------------------\\\\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\cdot \frac{\pm\sqrt{1-x^2}}{x}}{1-\left( \frac{\pm\sqrt{1-x^2}}{x} \right)^2}
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$\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies 
tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{\frac{x-1+x}{x}}
\\\\\\
tan(2\theta)=\cfrac{\pm2\sqrt{1-x^2}}{x}\cdot \cfrac{x}{2x-1}
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4 years ago