Can someone help me with some Geometry questions?
@sleepyjess
I'm not sure about the last part, but I know I can help with finding sin(x) and cos(y)
Okay :) Do you know SOH-CAH-TOA?
Yes, SOH = Sin(x) = opposite/hypotenuse CAH = cos(x) = adjacent/hypotenuse TAO = tan(x) = adjacent/opposite
Yep, so let's do sin(x) first, from angle x point of view: |dw:1430240723254:dw|
But we don't have the hypotenuse
Yep, so how do we find the hypotenuse?
Do we use tan?
Pythagorean Theorem :)
Oh... I;m still a bit confused :P
I kno how it is, but how do we apply it?
Whoops, sorry, I didn't get a notification that you responded We have a and b to use
Its all good :) I dont remember how to do the rest :P
\(8^2 + 6^2 = c^2\)
Oh, okeee... That would equal to 100
10^2
Yep, now \(\sqrt{100}\) = ?
10, yep haha
@sleepyjess The last part, X and Y are related because the must add up to 90 because each triangle angles add up to 180 and there is a right angle. 180-90=90. So X+Y=90
Yes, so we know that the hypotenuse is 10
ah, thanks @Phil1111 :)
And now we have to find sin?
Yep, now we find sin(x)
Alright, so, 6/10 = 0.6 Would that be the answer for sine?
Or am I missing something?
That's correct :)
So, that would also equal to 0.6
yep :)
Alright, this is what I have...
First, we have to find the hypotenuse by using the Pythagorean theorem. 8^2 + 6^2 = c^2 64 + 36 = 100 √100 = 10^2 The length of the hypotenuse is 10 Next, we have to find SOH. sin(x) = 6/10 sin(0.6) The Sine here is 0.6. Lastly, we have to find CAH. cos(x) = 6/10 cos(0.6) This means that the cosine also equals to 0.6. Does it sound good?
That is awesome :)
Yay! Now, for the last part of the question..?
Phil explained that one @sleepyjess The last part, X and Y are related because the must add up to 90 because each triangle angles add up to 180 and there is a right angle. 180-90=90. So X+Y=90
Alright, thnx so much!!!
Do you think you could help me in a couple more tho?
I have 2 more :P
As you were helping me with this one, I did the other 4 :P XD
I actually have to go get ready for gymnastics, but maybe @welshfella can help :)
Well, could you @welshfella? :P
ok
Alright, thnx so much!
I'll tag u, give me a sec
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