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ChaoticLlama
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
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.
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
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
BeautifulDreamer said: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:
G
Guest
Guest
that was the problem i needed help with, do i use tan 90 because it's a right angle?
ChaoticLlama
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
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: \(\displaystyle \,x \;\approx\;63.4^o\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
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?
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: \(\displaystyle \,x \;\approx\;63.4^o\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
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?