On line Study Group (1 Viewer)

[omigodwhenover]

Provided they write the latitude co-ordinates first, then you'd be right!

The general convention all over the world is to write latitude first (that's North or South) and then longitude (East or West), but who knows what tricks they might throw in!

Also ... arc length formula.

Arc length = @/360 x 2πr

where r is the radius of the circle and @ is the angle of the sector.

So using your example, if A is (23°S,127°E) and B is (23°S,145°E) then A and B lie along the 23°S line of latitude, a small circle. The angle of the sector is 145 – 127 = 18°.

Now, because it's a small circle, you'd need to be told the radius of that small circle. You won't have to work it out for yourself. It's 5891 km.

So distance = @/360 x 2πr
= 18/360 x 2π x 5891
= 1850 km

Hope this helps.

i was looking through some maths in society papers...cuz i ran out of other ones, and they give the formula for a small circle radius as r=Rxcosxangular difference between the two points, where the big R is 6400km (the radius of a great circle)

 

[PC]

Yes. Small circle radius, r = R cos @, where @ is basically the latitude.

It was always studied in the old MIS course, but I'm fairly sure that formula was always given in exams if it was needed. For the general course, it's not in the syllabus, so you won't need to know it for the exam.

It's a useful extension exercise for good classes, so I'd say that some classes would investigate it and have questions on it, but for the HSC, you'd be just given a small circle radius.

So ... don't sweat!

 

[Dr_Doom]

Yeah I found that in a text book the other day and never seen it before.. I ran up the line and the lady said you don't need to know it but you might as well.

 

[Dr_Doom]

18/20 then 16/20 then finally 22/22 xD

 

[emanuellasker]

Hey thanks Dr. Doom and omigodwhenover once again for your help. I was going to ask about ur solutions for the 1st question omigodwhenover, but I thought I'd be the gentleman.

 

[emily2005]

how else can u find the gradient of a straight line without using vx/hx?

 

[Dr_Doom]

how else can u find the gradient of a straight line without using vx/hx?

Select 2 coordinates then:

Change in Y
Change in X

 

[omigodwhenover]

Hey thanks Dr. Doom and omigodwhenover once again for your help. I was going to ask about ur solutions for the 1st question omigodwhenover, but I thought I'd be the gentleman.

haha yeah sorry bout that. thats exactly the kind of dumb mistake i make in exams when i dont concentrate properly, but then i actually went to do the question and i was like derrr ur a fool! HAha so yeah. Im pretty excited about maths cuz i cant really think of anything tricky they could possibly do that i havent seen before. I mean, it is general maths!!!

 

[Dr_Doom]

They do things in every question to trick you! It's sooo annoying!'
'

Also how do I do this:

The stopping distance of a car is proportional to the square of the car's speed. A car travelling at 60 km/h has a stopping distance of 40m.

If the stopping distance is 80m, what is the car's speed?

Spoiler

Answer: 85 km/h

 

[Dr_Doom]

oh wtf I just tried it again and got it right. !

40/60^2 = 0.0111111111

Square root of: 80/0.0111111111

= 84.85281374

= 85 km/h (nearest km/h)

 

[emanuellasker]

Hiya, I just had another question from the maths in focus textbook that I need help with. Here it is,
Goal posts are 3.5m apart. Greg stands 2.7m from one post and 3.2m from the other. Through what angle does he need to kick a football for it to go between the goal posts? (Answer to the nearest minute).

 

[omigodwhenover]

Hiya, I just had another question from the maths in focus textbook that I need help with. Here it is,
Goal posts are 3.5m apart. Greg stands 2.7m from one post and 3.2m from the other. Through what angle does he need to kick a football for it to go between the goal posts? (Answer to the nearest minute).

When u get a Q like this u can guarantee it involves trig, so the first step is to draw the triangle with sides 3.5, 2.7 and 3.2. The angle u are trying to find is the one opposite 3.5.
Next figure out which rule. Sine rule needs at least one angle, so u have to use cosine rule with the two sides next to the angle

 

[emanuellasker]

Thanks and good luck tommorrow!

 

[Dr_Doom]

Yeah good luck everyone xD Don't be nervous.. unless you have a 3 hour software exam afterwards like me...

 

[emanuellasker]

Just one mroe thing, Venn diagrams, what is their concept, I don't understand how to draw one even after reading the examples from my textbook? Can anyone plz help.
Ta,

 

[omigodwhenover]

ok basically u use them for intersecting sample spaces. The rectangle is the universal sample space and the circles represent the two sample spaces. So say there are 50 people, 37 like chocolate icecream and 41 like vanilla, so how many like both?
let A = those who like choc, B = those who like vanilla
AuB = universal subset
AnB = those who like both

AuB = n(A) + n(B) - AnB
50 = 37+41 - AnB
Therefore AnB = 28 people
then to find out those who only like chocolate simple subtract 28 = 9 people

 

[PC]

Venn diagrams are probably not required for tomorrow's exam. They're more of a way of solving some types of problems, but not the only way. Some schools will do questions which use Venn diagrams, but others won't.

Let's say you have a group of 300 students, and they can do some sports, say cricket or tennis. They could choose to do both or they could choose neither.

Let's say 100 people do both sports.
Let's say 200 people do cricket ... but since 100 people already do cricket AND tennis, we know 100 people do cricket only.
Let's say 130 people to tennis ... but since 100 people already do tennis AND cricket, we know 30 people do tennis only.

So we've worked out that:
100 people do both tennis and cricket
100 people do cricket only
30 people do tennis only
Now, since there are 300 people altogether, we've got another 70 that don't do any sport.

A Venn diagram is just a visual way of representing all that information. See we didn't need the diagram to work out the numbers. But it sometimes helps.

 

[emanuellasker]

Thnx, so PC you said they probably won't be in the exam anyway?

 

[emanuellasker]

Hey has anyone seen question 20 in the 2001 paper? How on earth do you do this.

 

[omigodwhenover]

Thnx, so PC you said they probably won't be in the exam anyway?

they wont ask u a specific question on it, but it can be useful in some situations like the one i mentioned. It is in the syllabus under the heading "suggested tasks" which means its not compulsory

As for question 20 in 2001 HSC, this question is almost exactly like the question i helped u with yesterday;
1. use trig to find BD (sin 76* = BD/100)
2. Use sine rule in triangle ABD (BD/sin25* = AD/sin 51*)
I usually aim to find as many angles in the triangles as i can first, and in this case u can find all angles before u start