Maths year 11
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Original post by z_o_e
You have two areas to find. One of the quarter of the circle, and one of the right-angled triangle. If you take the triangle's area away from the quarter circle's, you are left with the segment which is what you want.
The triangle's both sides are length 4.8 cm because they both are the radius of the circle. Can you visualise this?
(edited 7 years ago)
Original post by RDKGames
You have two areas to find. One of the quarter of the circle, and one of the right-angled triangle. If you take the triangle's area away from the quarter circle's, you are left with the segment which is what you want.
The triangle's both sides are length 4.8 cm because they both are the radius of the circle. Can you visualise this?
The triangle's both sides are length 4.8 cm because they both are the radius of the circle. Can you visualise this?
Nooo I can't :*(
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Original post by z_o_e
Okay imagine this:
You draw yourself a quarter of a circle of radius 1 as an example, what is the length of the two flat lines?
Now if you draw yourself a line between the two points where the two lines intersect the curve of the circle, you would get the exact same diagram as in the question.
Original post by RDKGames
Okay imagine this:
You draw yourself a quarter of a circle of radius 1 as an example, what is the length of the two flat lines?
Now if you draw yourself a line between the two points where the two lines intersect the curve of the circle, you would get the exact same diagram as in the question.
You draw yourself a quarter of a circle of radius 1 as an example, what is the length of the two flat lines?
Now if you draw yourself a line between the two points where the two lines intersect the curve of the circle, you would get the exact same diagram as in the question.
Radius of two flat lines is 2.
Yep I get it and then.
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Original post by z_o_e
Well... the length would be 1 each but let's see your working out on the question to see if there's any errors.
Original post by RDKGames
Well... the length would be 1 each but let's see your working out on the question to see if there's any errors.
But the thing is I don't know the triangles area as I don't know it's height
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Original post by z_o_e
Yes you do. The height is the same as it's length. As I said, they are BOTH the radius of the circle.
Original post by RDKGames
Yes you do. The height is the same as it's length. As I said, they are BOTH the radius of the circle.
Yepp
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Original post by z_o_e
Correct. Just write it to 3 s.f in order to get full marks.
Original post by RDKGames
Correct. Just write it to 3 s.f in order to get full marks.
Is this correct?
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Original post by z_o_e
Correct.
Original post by z_o_e
Have you been taught trigonometry? Not sure whether you have or not before I suggest what to do.
Original post by RDKGames
Have you been taught trigonometry? Not sure whether you have or not before I suggest what to do.
I love trigonometry.
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Original post by RDKGames
Okay. Here, do this. (Bonus points if you solve it using differentiation)
Ok ill give it a go when im back, I was hoping for a quicker question as im just about to get boozed up and vectors is my weakest topic. y u no pick integratioN? luv that stuff. Also even though this looks possible with a small bit of reading up again i didn't do further so not so good at in depth topic style question. Did you finish alevel this year?
Original post by z_o_e
Ah good, then this should be straight forward.
If you label the angle at the tower, lets say (called theta), then you know that is the opposite side divided by the adjacent side, with opposite being 4.5 and adjacent being 2.1 as you can see. You can find the actual angle by doing inverse tan. Of course, this angle is not the bearing because bearings are measured from the North, therefore you need to find the other angle long that line by doing , and that should give you the bearing. Don't forget bearings are shown as 3 digits.
(edited 7 years ago)
Original post by z_o_e
also youve probably done this question before by the looks of the vectors you gave but this is one on my fav integration cos it was funny watching the class go round in circles while trying to solve it ahah
∫ex2sin(ax)dx its meant to say
∫ e^x/2 sin(ax)
(edited 7 years ago)
Original post by TSRPAV
Ok ill give it a go when im back, I was hoping for a quicker question as im just about to get boozed up and vectors is my weakest topic. y u no pick integratioN? luv that stuff. Also even though this looks possible with a small bit of reading up again i didn't do further so not so good at in depth topic style question. Did you finish alevel this year?
No further? Hah, good luck mate, I'll give you a medal if you actually solve it then. And yes, I've finished A-Levels this year.
Original post by TSRPAV
also youve probably done this question before by the looks of the vectors you gave but this is one on my fav integration cos it was funny watching the class go round in circles while trying to solve it ahah
∫ex2sin(ax)dx its meant to say e^x/2
∫ex2sin(ax)dx its meant to say e^x/2
You might not want quote a Y10 student for that...
(edited 7 years ago)
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