# A Level Maths Trigonometry

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#1
Hi Guys,

I need help with a question please:

I've done part a and got 1.194 x 10^-5 however it's the next part that's bothering me.

Now I do get that its talking about the arc length but I dont get what I have to do with the cos small approximation angle info?? Can I not just simply use the angle in part a and the radius which is the distance??
Last edited by Cos00088; 3 months ago
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3 months ago
#2
They want a) to 5 sig fig, but the value is approximately correct.

For b) they obviously want you to approximate the arc with a straight line (chord), so the sector becomes an isosceles triangle. However, sin would seem to be the obvious trig function to use (historically, it is related to the length of a chord), However, for small angles, this is simply the arc length.

Where does it come from?
Last edited by mqb2766; 3 months ago
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#3
Thank you soo much

So I use the cos rule to get the length of the chord. However the angle would be:

1 - 1/2x^2 = (where x will be the angle calculated in part a)

Right?
Last edited by Cos00088; 3 months ago
0
3 months ago
#4
(Original post by Cos00088)
Thank you soo much

So I use the cos rule to get the length of the chord. However the angle would be:

1 - 1/2x^2 = (where x will be the angle calculated in part a)

Right?
Can you sketch what youre trying to calculate and upload?
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#5
Thanks again for helping me out

Last edited by Cos00088; 3 months ago
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3 months ago
#6
(Original post by Cos00088)
Thanks again for helping me out

You can use the cos rule like that, though squaring lengths (x's) will be large (small).

The obvious thing to do would be to split the isosceles triange into 2 and use sin() on the half angle (half chord distance), but using the sin() local approximation is just the arc length?

Where is the question from?
Last edited by mqb2766; 3 months ago
0
#7
(Original post by mqb2766)
You can use the cos rule like that, though squaring lengths (x's) will be large (small).

The obvious thing to do would be to split the isosceles triange into 2 and use sin() on the half angle (half chord distance), but using the sin() local approximation is just the arc length?

Where is the question from?
My teacher gave it to me so I've got no idea. I would've asked her but shes on leave.

I
Last edited by Cos00088; 3 months ago
0
3 months ago
#8
(Original post by Cos00088)
My teacher gave it to me so I've got no idea. I would've asked her but shes on leave
Id just ask her when she gets back.
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#9
(Original post by mqb2766)
Id just ask her when she gets back.
I've got exams next week and shes not back till after half term but thnx for helping I've got an idea as to what I should do 0
3 months ago
#10
(Original post by Cos00088)
I've got exams next week and shes not back till after half term but thnx for helping I've got an idea as to what I should do You can use the cos rule, but youre subtacting 10^-10 from 1 then multiplying it by ~10^12. Its easy to get numerical problems. The obvious thing to do is the sin rule, but thats not worth 5 marks and is "no" different from calculating the arc length using the angle directly.
Last edited by mqb2766; 3 months ago
0
#11
(Original post by mqb2766)
You can use the cos rule, but youre subtacting 10^-10 from 1 then multiplying it by ~10^12. Its easy to get numerical problems. The obvious thing to do is the sin rule, but thats not worth 5 marks and is "no" different from calculating the arc length using the angle directly.
Okay, I've got it now! Thank you soo much
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