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Mathematics 9 Online
OpenStudy (ohmybookness):

Did I solve these two questions correctly so far? I'm having trouble solving #2

OpenStudy (ohmybookness):

OpenStudy (ohmybookness):

@Nnesha Can you check for me?

OpenStudy (ohmybookness):

@Nnesha

OpenStudy (ohmybookness):

I confidently got 8.6 for my first method. Now I don't know what other method to use :/

Nnesha (nnesha):

hmmm. are you in school right now ?

OpenStudy (mathmale):

Do I see "Graded Assignment" on this worksheet? Won't check your work, but will answer specific questions about #2.

OpenStudy (ohmybookness):

No this is just HW.

OpenStudy (mathmale):

Questions?

OpenStudy (ohmybookness):

What other method should I use to solve this problem

OpenStudy (ohmybookness):

I did the Tan55=15/opp but I'm not sure if that's the method I should use.

Nnesha (nnesha):

hmm ok. you skipped one step for 1st question. how did you get pi/6 ??

OpenStudy (mathmale):

To be honest, your worksheet is hard to read. I can just barely make out the instructions. To find the length of the shorter side, you could use the Pyth. Theorem, or continue to use trig. Specific questions about that?

OpenStudy (ohmybookness):

@Nnesha from 17pi/6

OpenStudy (ohmybookness):

@mathmale what numbers should I plug in?

OpenStudy (mathmale):

Please provide structure (equations) and specify what your objective is.

OpenStudy (ohmybookness):

What I have so far is tan55=15/opp

OpenStudy (ohmybookness):

Is that the correct equation to use to solve the shorter leg of triangle ABC

OpenStudy (mathmale):

AB is your hypotenuse. The longer leg (not the hyp) is of length 15. You want the shorter side? Personally I'd write out\[\tan 35=\frac{ opp.side }{ adj.side }= \frac{ x }{ 15 }\]

OpenStudy (mathmale):

Please go over my suggestion. Does it make sense to you? If not, ask questions. if wrong, say so.

OpenStudy (ohmybookness):

the hypotenuse is not 15?

OpenStudy (ohmybookness):

Wow that is very helpful!!

OpenStudy (mathmale):

No, it's not. The hypo is the longest side. That 15 denotes the length of the longer leg, not the length of the hypot.

OpenStudy (mathmale):

My great pleasure.

OpenStudy (ohmybookness):

Where would I go from there? do I divide? I know tan 35 =.7

OpenStudy (ohmybookness):

So my answer for a is wrong. How would I find the length of the hypotenuse?

OpenStudy (mathmale):

as before, tan 35 deg = opp / adj ...which you could (if you wish) solve for the unknown, which is the opp side. Skipping that step, we get 0.700 = opp / 15. Mult both sides of this equation by 15 to clear out the fraction. Write your answer as follows: opp = ( ? ) (0.700) = value?

OpenStudy (ohmybookness):

10.5

OpenStudy (ohmybookness):

Do you know what I did wrong on my first method?

OpenStudy (mathmale):

I haven't done the math, but that looks very reasonable. If you want to check your work, find the length of the hypo. using the Pyth. Thm. and then determine whether (length of hyp) * sin 35 deg gives you the same 10.5 result.

OpenStudy (ohmybookness):

I solved using sin(theta)

OpenStudy (mathmale):

Again, the material is hard to read, so I'm not spending much time with it. The major fault by far is that you considered the side of length 15 to be the hypotenuse. That side is not the hyp.; rather, it's the longer leg.

OpenStudy (ohmybookness):

Can I write it out for you? Maybe it'b be easier. Trig is very hard for me to grasp.

OpenStudy (mathmale):

There's more than one way in which to skin a cat. You could use 35 deg, 55 deg, theta, whatever, but you must recognize that the side labeled 15 is not the hyp. Can you adjust your cell phone camera to produce a brighter image with more contrast?

OpenStudy (ohmybookness):

How do I solve for the hypotenuse. I can get everything else, I just assumed 15 was the answer.

OpenStudy (ohmybookness):

|dw:1456543444364:dw|

OpenStudy (ohmybookness):

c2=a2+bc?

OpenStudy (mathmale):

Between the 2 legs of your triangle you see a box:|dw:1456543482539:dw|

OpenStudy (mathmale):

yes, (opp)^2 + (15)^2= (hyp)^2.

OpenStudy (ohmybookness):

How do I solve for the hypotenuse not knowing what the opp side is (even though I do)

OpenStudy (mathmale):

You'd use the cosine function. This says: cos 35 deg = adj/ hyp Here, your adj side has length 15. Thus, cos 35 = 15/hyp ... this can be inverted to obtain \[\frac{ 1 }{ \cos 35 }=\sec 35= \frac{ hyp }{ 15 }, or 15 \sec 35 = hyp\]

OpenStudy (mathmale):

Your calculator does not have "sec" so you'll have to stick with the cosine function. 1 / cos 35 = 1 / 0.8192 = hyp / 15

OpenStudy (mathmale):

So 15/cos 35 = hyp or 15 / 0.8192 = hyp

OpenStudy (ohmybookness):

18.3?

OpenStudy (mathmale):

I'm not checking this math, but would readily say your answer looks appropriate. It's larger than 15 and larger than 10.5, and thus suits the longest side.

OpenStudy (ohmybookness):

Thank you so much!!!

OpenStudy (mathmale):

If you want to be sure, check whether (18.3)^2 = (15)^2 + (10.5)^2.

OpenStudy (mathmale):

My great pleasure; appreciate your interest and perseverance. Best wishes!

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