tangent problem

[Guest]

 

[ChaoticLlama]

It's and easy question if you know what to do.

Think about it this way, when you square a number, what is the reverse operation? A square root.
Hence, \(\displaystyle \begin{array}{l}
8^2 = 64 \\
\sqrt {64} = 8 \\
\end{array}\)

In this case, you want to solve for x. So consider this, what is the reverse operation for trig functions? Their inverse functions.
\(\displaystyle \begin{array}{l}
\tan (x) = \frac{7}{{11}} \\
x = \tan ^{ - 1} \frac{7}{{11}} \\
\end{array}\)

It's almost purely a calculator problem, unless you are dealing with special angles.

 

[ChaoticLlama]

tan90 = 3 radical 5 over 3

you have apparently changed the exact question in the time I took to respond. And are you sure the question is tan(90)? The tan function is undefined for all multiples of 90 degrees.

EDIT: You have changed your question about four times over the interval of which I have responded to your topic. :evil:

 

[Guest]

that was the problem i needed help with, do i use tan 90 because it's a right angle?

 

[ChaoticLlama]

tan(90) is and undefined expression. It means nothing.

Go here to this website

 

[soroban]

Hello, Dreamer!

If that is the correct problem (sides $\displaystyle 3$ and $\displaystyle 3\sqrt{5}$)
$\displaystyle \;\;$and the problem is to find angle $\displaystyle x$ in degrees . . .
then we can proceed.

Relative to angle $\displaystyle x$, the adjacent side is $\displaystyle 3$ and the hypotenuse is $\displaystyle 3\sqrt{5}$.

"Adjacent" and "hypotenuse" should suggest the cosine function . . . right?

So we have: $\displaystyle \,\cos(x^o) \:=\:\frac{3}{3\sqrt{5}}\;=\;\frac{1}{\sqrt{5}}$

Then: $\displaystyle \,x^o\;=\;\cos^{^{-1}}\left(\frac{1}{\sqrt{5}}\right) \;= \;63.43494882$

Therefore: $\displaystyle \,x \;\approx\;63.4^o$


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Why are you playing with the $\displaystyle 90^o$ angle?

All right triangles have a $\displaystyle 90^o$ angle.
$\displaystyle \;\;$Does that mean that the answer will always be the same?