a few questions i can't do (1 Viewer)

[Grey Council]

From the cambridge year 11 textbook. sorry, won't type up questions, need to sleep and its getting late. If no one has replied, or someone asks, i'll type up the questrions tommorow afternoon.

Q 21, exercise 14B, part 2.
wtf, how do you do this, i can't seem to get the answer. :-\ It looks easy, i'm prolly making a stoopid mistake. meh

Q 23, exercise 14D
huh? According to my diagram, this isn't even possible.

more to come (hopefully not, but its cambridge, and with its reputation, i'll prolly be asking more. hehe, first time using cambridge )

 

[freaking_out]

by not typing it up- ur eliminating any chance of an 03'er answering the question for u.

 

[KeypadSDM]

I.e. Us

 

[freaking_out]

Originally posted by KeypadSDM
I.e. Us

yep, and also the 1337's like turtle and stuff (i know this is the 3u forum- but still! )

 

[Grey Council]

hrmph

hrmph, not happy. Had to draw graphs.

q 21, 14B
The diameters of two circular pulleys are 6cm and 12 cm, and their centres are 10 cm apart.

Find, in cm correct to one decimal place, the length of a taut belt that goes round the two pulleys.

q 23, 14 D
The diagram shows two rectangles. Each rectangle is 6cm long and 3cm wide, and they share a common diagonal PQ. Show that tan theta = 3/4

obviously the diagram is not to scale, but you can see what the book has drawn.

 

[LadyMoon]

The second one:
(Refer to the altered pic)

A= B + O

tan(O)= Tan(A) - tan(B)

Tan A=6/3 =2.

Tan B=1/2

Tan O= Tan A - Tan B
Applying the tan formulae,

Tan O= Tan A - Tan B/(1+tanAtanB)

Therefore: Tan O=2-1/2/(1+1)
TanO=3/4.

 

[Grey Council]

hrm, referring to your picture:
tan(O) = opposite over adjacent = 3/4
however, adjacent is 3, not 4.

wtf, that confused the shit outta me. lol

but i see how its done

 

[wogboy]

q 21, 14B
The diameters of two circular pulleys are 6cm and 12 cm, and their centres are 10 cm apart.

Find, in cm correct to one decimal place, the length of a taut belt that goes round the two pulleys.

The best way of attacking this question is with a geometrical approach. Have a look at my attached diagram. O and P are the centres of the circles.

angle ORQ = angle RQP = 90 deg (tangents to the radii)
OR = 3cm (data)
PQ = 6cm (data)
OP = 10cm (data)
angle OSP = 90 deg (data)
angle OSQ = 90 deg (supplementary angles)
so OSQR is a rectangle
so SQ = 3 cm
so SP = 3cm

therefore OS = sqrt(91)cm (Pythagoras)
QR = sqrt(91) cm

you can assume by symmetry that the green line in my diagram (i.e. QR) is equal in length to the red line. So the total length of the straight parts is 2*sqrt(91) cm. Now to find the curved parts.

angle OPQ = arccos(0.3)

So the angle at which the band goes around the large wheel is 360 - 2*arccos(0.3). This makes the curved part of the band on the long wheel:

[360 - 2*arccos(0.3)]*pi/180 * 6 cm
~ 22.51 cm

Using similar methods, you can find the curved part of the band on the small wheel to be (I'm too lazy to explain in detail):

{360 - 2*[180 - arccos(0.3)]}*pi/180 * 3 cm
~7.60 cm

Add all these lengths and you get:
~ 49.2 cm

 

[LadyMoon]

Originally posted by Grey Council
hrm, referring to your picture:
tan(O) = opposite over adjacent = 3/4
however, adjacent is 3, not 4.

Oh yeah...
Ok now thats "confusing the shit outta me" too.

Can someone else explain why it is so? Surely the answer can not be incorrect, what other way is there of doing?

 

[wogboy]

hrm, referring to your picture:
tan(O) = opposite over adjacent = 3/4
however, adjacent is 3, not 4.

Ladymoon's solution is correct. The reason why that's so is because the opposite is 9/4 units long (while the adjacent is of course 3 units long), hence tan(O) opposite over adjacent = (9/4)/3 = 3/4. It's just the ratio of the opposite to the adjacent that matters, not the absolute lengths of the sides.

There's nothing to worry about !!

 

[Grey Council]

oh, right. lol, i was worried for quite a while. ^_^
Thanks wogboy and Ladymoon.

And as for the taut pulley question, I did the EXACT same thing you did. And I couldn't get 49.2. blah, i prolly made some silly mistake somewhere. hehe, i'll give it another shot tommorow.

 

[LadyMoon]

Originally posted by wogboy
Ladymoon's solution is correct. The reason why that's so is because the opposite is 9/4 units long (while the adjacent is of course 3 units long), hence tan(O) opposite over adjacent = (9/4)/3 = 3/4. It's just the ratio of the opposite to the adjacent that matters, not the absolute lengths of the sides.

There's nothing to worry about !!

Thank you!
I was freaked for a second....

 

[Grey Council]

hah! I was freaked for like half an hour. I tried changing theta around, thinking that it was a typo and for tan theta to equal 3/4 it would have to be on the OTHER side of the rectangle, but I just got more confused after that. Then I freaked. lol, i didn't even attempt the question, I thought since my diagram was obviously wrong, whats the point of continuing. hehe

damn, wogboy is smooth. At least he helps our heaps. ^_^ AND he is one of the only people I know who uses the notation arccos, arctan, arcsin. heheh

 

[wogboy]

damn, wogboy is smooth.

heh, as per my avatar

he is one of the only people I know who uses the notation arccos, arctan, arcsin. heheh

I picked that up at uni, my lecturer and tutors said that they prefer people to write arcsin instead of sin^-1 etc (they won't make you lose any marks over it, but it may tempt them to be a little more generous with marks if you don't get the correct answer ). And yes I do get a bit annoyed when people write sin^-1 instead of arcsin.

 

[freaking_out]

Originally posted by wogboy
heh, as per my avatar ...

i liked your old avatar (when u had wogboy there).

 

[Grey Council]

thats why I said smooth. hehe

and yes, arcsin is so much easier to understand, esp on these forums. blah, if you write sin^2x people get confused if its to the power of two x, or whatnot.

on the other hand, people usually don't know arcsin either. but arcsin is easier to write.

 

[KeypadSDM]

Alas, it's just people not writing it correctly which causes confusion. They try to make it shorter, and fail to use the suggested notation.

Sin^2[x] isn't ambiguous, since it uses the notation correctly.