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Mathematics 17 Online
OpenStudy (gabby_taylor1224):

Find the measure of angle x. Round your answer to the nearest hundredth. (please type the numerical answer only)

OpenStudy (gabby_taylor1224):

OpenStudy (johnweldon1993):

hmm, which class is this for?

OpenStudy (gabby_taylor1224):

Geometry

OpenStudy (johnweldon1993):

So lets see, have you used trig before? like sin, cos?

OpenStudy (gabby_taylor1224):

a little bit

OpenStudy (johnweldon1993):

Okay good, that makes it easier then...and it's very easy I'll explain step by step

OpenStudy (johnweldon1993):

So we have a triangle |dw:1445434069149:dw|

OpenStudy (johnweldon1993):

Notice that we have an angle that we want....as well as 2 sides of the triangle known In reference to the angle...we know the side opposite that angle.....and we also know the side adjacent to the angle...right?

OpenStudy (gabby_taylor1224):

yes

OpenStudy (johnweldon1993):

So think back to your trig functions sine cosine tangent Which one of those uses those sides? the opposite and adjacent?

OpenStudy (gabby_taylor1224):

cosine?

OpenStudy (johnweldon1993):

Not quite There is a VERY easy way to remember which functions require which sides Remember SOH-CAH-TOA S sine O opposite H hypotenuse C cosine A adjacent H hypotenuse T tangent O opposite A adjacent So sin = opposite / hypotenuse cos = adjacent / hypotenuse tan = opposite / adjacent

OpenStudy (gabby_taylor1224):

Ohh okay

OpenStudy (johnweldon1993):

Now I know you're like wtf why would that help? how will I remember that? turn it into a sentence Like Some - Old - Horse - Caught - Another - Horse - Tripping - Over - Apples Haha or something corny like that XD

OpenStudy (johnweldon1993):

So now you see that \[\large \tan(x) = \frac{opposite}{adjacent} = \frac{8}{15}\] right? now there is only 1 more step

OpenStudy (gabby_taylor1224):

okay, I'm starting to understand

OpenStudy (johnweldon1993):

So since we have \[\large \tan(x) = \frac{8}{15}\] We need to solve for 'x' *notice we are solving for tan(x) there* how would we solve for just 'x'? Well if you've noticed...every trig function (sin, cos, tan) also has an inverse function (arcsin, arccos, arctan) And what we see is that if we take the inverse function, of the function...they cancel i.e. \[\large arctan(tan(x)) = x\] So it seems with our \(\large tan(x) = \frac{8}{15}\) if we just take the arctan of both sides, we can solve for 'x' \[\large x = arctan(\frac{8}{15})\] Plugging that into any calculator **************in degrees mode****************** will give you your angle :)

OpenStudy (gabby_taylor1224):

so would my answer be 2.128

OpenStudy (johnweldon1993):

Not quite That might be in radians *that's why I tried to emphasize that the calculator has to be in degrees mode before you do the calculation Because what I get is \[\large \arctan(\frac{8}{15}) = 28.07 \text{ Degrees}\]

OpenStudy (gabby_taylor1224):

okay, thank you!

OpenStudy (johnweldon1993):

No problem! I hope it made sense :)

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