#cosine-sine and tangen
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Considering you know two sides, you can just use Pythagoras' theorem.
Ok
So 8.999cm right?
No. Obviously, the side can't be greater than the hypotenuse.
I don’t know can you solve it I can’t find any other answer than this
You just didn't apply it correctly. We have:
AB^2 + BC^2 = AC^2
So:
AB = √(AC^2 - BC^2)
5cm?
Has to be
Uh, no...
How did you get that?
Idk do you know the answer?
I didn’t write but I used Kat+kat=hyp it’s wrong but I need the answer
Why just add them?
As I told you above, we use Pythagoras' theorem:
AB^2 + BC^2 = AC^2
From here AB = √(AC^2 - BC^2). So, calculate it.
2.9816103031751148012063558405269
Well, no need for so many decimals.
We are given numbers with two significant digits, so we also leave two. Thus, AB ≈ 3.0 cm.
Yes.
Well, we know an angle and an opposite side, and we need to find the adjacent side and the hypotenuse.
Which trigonometric functions should we use?
Cos right?
Sin (29.7) = 4/hypotenuse i don’t know how I use this
No.
Is it 8cm?
Well, we have sin(F) = ED/FD. So, you can express FD from here.
Well, FD is closer to 8.1 cm.
So the other one is 7,04cm
Well, closer to 7.01 cm. Also, we should probably keep the same amount of significant digits, so just 7.0 cm.