Ez triangle

[latent kelp]

well you need to take the cos of θ

[pastel sleet]

0.54

[carmine cave]

The arccos/inverse cosine to be exact.
$\newline \cos{\theta} = \frac{6}{11}\newline\theta = \arccos{\left(\frac{6}{11}\right)}$

[dense hill]

0x

[carmine cave]

And make sure that your calculator is set to degrees.

[pastel sleet]

So that gives me the opposite side but how do I find the theta

[carmine cave]

No, it gives you θ. Look at the equation.

[carmine cave]

The cosine of an angle is equal the length of the adjacent side divided by the hypothenuse.

[pastel sleet]

So 56.9 right..?

[carmine cave]

Precisely!

[pastel sleet]

Bet thnx

[pastel sleet]

Which of the following is possible?
a) csc theta = -5
c) tan theta = pi
b) sin theta = 19/18
d) sec theta = 0

[carmine cave]

Do you know how each of those trigonometric functions is defined?

[pastel sleet]

No

[carmine cave]

$\newline\csc{\theta} = \frac{1}{\sin{\theta}}\newline$
$\newline\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}\newline$
$\newline\sec{\theta} = \frac{1}{\cos{\theta}}\newline$

Knowing these, and knowing that $|\sin{\theta}|$ and $|\cos{\theta}|$ can never be bigger than 1, what does that mean for each of these functions in terms of possible solutions?

[dense hill]

0x

[pastel sleet]

So it could be A and D

[carmine cave]

Not quite. A is correct but D can never be zero. Can you see why?

[pastel sleet]

Cause sec can’t be 0 cause it’s dividing by 1?

[carmine cave]

Well the numerator can never be zero which means that the whole thing can't be either.

One more of these is possible though.

Can you find it?

[pastel sleet]

no..

[carmine cave]

What happens to tan(θ) as sin gets larger (and cosine therefore gets smaller)?