sidereal day

[mollykins]

Because the earths orbit is eliptical, wouldn't that mean that a sidereal day varies aswell as a solar day?

[factor-fiction]

This may help
http://en.wikipedia.org/wiki/Sidereal_time

[mollykins]

I tend not to trust wikipedia . . . .

hi factor how are you? my english exam today was easier than maths yesterday.

[mollykins]

aaahhh, right at the bottom it says that it does vary, i was wondering that but don't think it'll be in the exam. But can it never be as long as a solar day?

[factor-fiction]

Hope you remembered to use apostrophes in English.
Based on what you said about your Maths and your experience with revision questions I think your most likely grade is A. To get A* you need to be able to handle questions which are a little out of the ordinary such as the one involving several interlinked/simultaneous equations.
What's next?
Are you doing French- you should be pretty good at that with your family background?

[mollykins]

I'm aweful at french and haven't done it for 2 years.

In the exam, there was a table which created 9 simultaneous equations (all addition) with three variables which you had to work out, what made it worse was that one was a negative number.

[factor-fiction]

(Spelling is AWFUL)
I'll have a look at the exam question and post an answer for you -maybe Friday.

[mollykins]

I answered it ok thanks, but there was a nasty triganometry question which i thaught was impossible.

[mrs.chappie]

Hey mollykins, you're a similar colour to factor30! (Totally irrelevant .... just thought I'd throw it in the mix).

[:o)

[factor-fiction]

I'll look at the nasty trigonometry question too for you. Did you need to use the sine rule or cosine rule? What was nasty about it? (It's usually only bearings that make them nasty)

[society]

Molly, the saying “practice makes perfect” is a very true. I suggest you use your posting opportunities on here to practice you writing, i.e., your grammar, spelling, sentence structure etc. Just a thought, not criticising...

[mollykins]

it was something like; a man is 1.8metres tall looking at the top of a hill at about 66* creating a triangle, with the base at something like 1850 metres away, plus there was a mass that was a certain height (i think about 300 metres), on top of the hill. and you had to work out the total height of the mast above the ground.

[mollykins]

*mast.

[mollykins]

You're given that the sin or cos of the angle is 0.5 but you need to use tan!

[factor-fiction]

Easy. You should know
Sin 30 =0.5 and Cos 30= √3/2
So tan30=sin30/cos30=1/√3 =√3/3

[mollykins]

We've never been taught any of those . . . .noone i talked to from my maths group knew how to do it.

[mollykins]

I'm wondering how much more we've not been taught, and these are the biggest and hardest maths exams.

[mollykins]

I think i divided 1850 by the 0.5 and added the 300 and 1.8 . . . it was the only thing I could think of.

[mollykins]

I mean times it, so it would be halfed, then added 1851.8.

[mollykins]

oops again, then added the 301.8

[factor-fiction]

I'm surprised you haven't covered the main sin values (sin 0=0, sin 30=0.5, sin 60=√3/2 , sin 90=1)
You can work it out even if you don't know it.
Sketch a right angled triangle with a side at an angle x from the base. if sinx=0.5 then you know OPPOSITE /HYPOTENUSE = 0.5. So let your triangle have a height of 0.5 and a hypotenuse of length 1. Call the base length b
Use pythagoras: b² + (0.5)²= 1²
So b²=1-¼ =¾
so b=√3/2
So tan b= opp/adj= =0.5/(√3/2)= 1/√3

Fiddly, I know if, you aren't used to it, but that is what's required for A*