Find x A. 9 B. 18 C. 9√3 D. 3√3
The equation?
We will be using Tangent. x is the hypotenuse and 9 is the opposite. 30 is the theta: \[tan30 = \frac {9}{x}\]
\[tan30 \approx 0.577\]
\[0.577 \approx \frac {9}{x}\] Multiply x on both sides. then divide tan30 on both sides and solve. \[(0.577)(x) \approx 9\] \[x \approx \frac {9}{0.577}\] \[x \approx 15.59\]
Hmm, I think I did something wrong.
haha thanks for helping me tho. I am having trouble with this unit
do you have any other idea on how to do this?
Yeah I'm kind of confused as well. >.<
haha ok thanks anyways. Do you know of anyone else who can help?
@thomaster
OOOOH I see what I did wrong. I needed to use sine and not tangent.
sin30 = 0.5 9/0.5 = 18
Ha, silly me. :P
haha thank you so much!
can you help with one more?
sure
Using SOH CAH TOA, do you know which triginomic ratio we need to use? Cosine, sine or tangent? |dw:1394823567997:dw|
wow this is making me see how truly bad i am at this... im not really sure... promise im not just tryin to get answers haha
oh and the options are A.9
B. 9√2 C.4.5√2 D. 18√2
SOH means: sine = opossite/hypotenuse CAH means: cosine = adjacent/hypotenuse TOA means: tangent = opposite/adjacent They are abbreviations. We have the opposite measurement and are looking for the adjacent. So which ratio do we need to use?
TOA?
Yes. We are going to use the tangent. So: \[tan45 = \frac {18}{x}\] Do you know how to find x?
ugh. no not quite
Well we learned in previous math lessons that to find a missing variable you need to isolate the unknown variable. So in this case we need to find a way to move the "x" so that it is by itself. Are you understanding so far?
ya i understand what you r saying
But we need to know what tan45 is. But we can't do that calculation so quickly, which is why we have calculators. Do you know how to find tan45 with a calculator? :3
no
Some calculators are different. Do you have one at home?
ya
Ok, are there tan, cos and sin buttons on it?
is it 1? i put 45 tan
Yep!
yay i actually got something right! haha
So now we have: \[1 = \frac {9}{x}\] So now let's isolate the x. What's the first step?
multiplying 9 and 1?
Nope, because x is the denominator then we must multiply x on both sides. That gives us 1x on the left, but what will happen on the right?
oh ok..... 9x? i have no idea
Since it's already being divided by x, it will be canceled out by the multiplication. So now we have: \[1x = 9\] What is the last step?
oh ok i see.
umm devide 1 and 9 so its 9
Yes. So x = 9. And to check, we have to make sure that tan45 = 9/9 tan45 = 1 x = 9 1 = 9/9 1 = 1 :D
yay thank you so much!!!
Aaaah I mixed up 18 with 9 somehow. .-.
Hold on xD
1 = 18/x 1x = 18 x = 18/1 x = 18 1 = 18/18 1 = 1 Yeah xD
so i was right with 9?
Omg I'm so stupid 18 was the hypotenuse .-.
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