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hi Bob
I reckon i have it about 98% now
just a little confusion arising out of the difference in approach between the activity text and the way/s in which you dealt with the various aspects.
Many thanks for your patient help...........
best wishes
David/diamond
re casio etc.......
i should have written polar coordinates rather than bearings...
does anyone have either of the follow casio scientific calculators:
fx-83GT Plus or
fx-85GT Plus ?
I believe that in operation they are identical. The difference between them is the power source - the "85" having a different battery and solar panel.
Anyway - i would like someone to explain to me how to calculate polar bearings using either of the said calculators ..............
hi Bob
well, it is "tomorrow" 02:40 Thursday
i woke around an hour ago and thought i saw a fair amount of light -
which led me to "being with Jade" and with the aid of a circle with 15 degree divisions i deduced approximate bearings from all points.
my next step is to calculate them all again and compare my calculations with my "deductions"
the only points - bearings which are in doubt now - i think - are for points G, H and J
so i will do this first (after breakfast)
after that i will want to calculate all of the others for my own peace of mind.
watch this space .................
regards thanks etc
David
noted bob
i think i am aware of most of what the text says such as about which quadrant applies but there is something or there are some implications of some of the things in the text of which i am unaware (not blissfully i might add!)
Well i will come back to it tomorrow
david
bob
i have not cracked it -
this is just a hunch/feeling but from what you have said recently i have deduced that the bearing from D to O may be 045 degrees
and that
the bearing from F to O may be 233.2 degrees?
are these bearings, at least, correct?
hi bob
i am trying to do what you suggest but, so far i can only see:
D back to the start at O is up and right from quadrant 3 to point O
so D is in Q3 -x and -y territory
the text of part two says to subtract the angle from 270 degrees for certain points including point D (i think that this does not necessarily apply when following your advice)
point F is in Q1 F-O direction is down and left
Now i will try to focus on how to arrive at the bearing from the angle, taking into account all that you have said and what i am re-reading from the activity text.
watch this space .............
david
hi bob
well i followed the instruction book for my calculator to program it for polar coords.
it finished with r=2,bearing=45 and this was displayed on the screen.
i was very unsure of what to do next - i had the feeling that another key or perhaps two were needed to , as it were, finish the process, but nothing else was indicated in the book.
so
i have, instead of proceeding with that, calculated the "rectangular coords for points F-J. that is to say the first part before the subtraction or addition to arrive at the three figure bearing.
below are my "first part figures:
from point F atan 3/4 (0.75) = 36.8 degrees
" " G " 4/4 (1) = 45 "
" " H " 4/4 (1) = 45 "
" " I " 5/4 (1.25)= 51.3 "
" " J " 5/6 (0.83)=39.7 "
If these figures are correct i will next do the conversion calculations to convert these to polar coords. (I will get back to doing the polar calcs on the casio .......)
I need a cup of tea now ...........
david
thank you Bob
that one is a break through
I recently saw a tv series by a prof. marcus de sautay (what a name) prof. of maths at i think oxford uni. I found the series very interesting and there i heard about Rene Descartes and others.
So
am i right in arriving at 225 degrees as the bearing from point D (270 - 45) ?
I think i will now play with my calculator and see if i can get it to give me the polar coordinates for points F - J inclusive. When i have the results i will "walk" them past you .......
All of this is keeping me occupied..........
till next time
David
hi Bob
thanks for your latest - I have made my own copy
what i really need to do now is to calculate the bearing for point C using the right angled triangle method.
I believe your figure was 315 degrees which seems to indicate that with my chosen method 315 would be arrived at from 360 - 45 degrees = 315 degrees?
But if that is so - i cannot see from the calculation given in the text for point B
(1/2 = 0.5 atan (0.5) = 26.6 degrees) how i do the equivalent calculation for point C
That is my current sticking point (there i go again with my unintended puns)
My current thinking is that i would like to calculate three figure bearings for all of the points using cartesian coordinates and then convert them all to polar coordinates in the way that the text describes.
Am i right in thinking that "cartesian" is effectively the same as "rectangular" (as opposed to polar) in the context of coordinates? What i mean to say is rectangular in the sense of being calculated using right angled triangles.
I hope i am making sense and transferring my thoughts effectively.
This could run and run as they say ...............
David
thanks for that bob
i live in Stafford - the county town
my next step will be to calculate the column for value of the angle (can't find the proper sign on the keyboard) - but only down to point E - using the right angled triangle method shown in the part two text.
At the moment i am unsure of how to do this - despite having done all of the triangle stuff on maths is fun.
till next time .............
david
Hi Bob
just after 7 am here in uk
a quick question
re the five column table in part two
column two (from the left) is headed "value of r" - r being one of the polar coordinates
it seems pretty obvious but i need to be certain .....are all the true figures for this column as the ones in the table from part one headed "distance (in a straight line) from O"?
david
Hi again Bob
Now that I have digested at least some of what you have told me, I have been able to clear up somewhat and can now turn my attention to what remains to be clarified in my mind.
The part one figures of mine seem to be ok. I did proceed on the assumption that the point C figure of four was a typo.
I am trying to follow the text itself (of the activity) and the given examples, figures and methods described.
Also noting relevant information such as how to calculate the three figure bearings and the fact that the text indicates to only use polar coordinates for points F to J. (and to use right angled triangles such as the one in the text labelled "OAB" for the points up to E only - this is how i attempted to arrive at three figure bearing figures (non-polar) for points up to and including point E.)
Perhaps this time around you could check my coordinates copied from the text as:
X Y Point
0, 0 O
0, 1 A
2, 1 B
2, -2 C
-2, -2 D
-2, 3 E
4, 3 F
4, -4 G
-4, -4 H
-4, 5 I
6, 5 J
Meanwhile i will try again with the three figure bearing figures back to O from points C, D and E. I note that
your figure for C was 315 degrees. Also what you wrote about the sequence.
From the square root figures for distances back to zero I calculate the following kilometer figures back to zero from:
point B 2.24 km point C 2.83 km point D as point C 2.83 km point E 3.61 km point F 5.00 point G and point H 5.66 km point I 6.40 km and, finally, point J 7.81 km. Do these figures look right to you?
many thanks for "bearing" with me. (no pun intended...)
hallo Bob
thanks for your response
I think I have done too much without the assurance that what i have done to date is correct.
I will give below the figures that i have arrived at for the distances (part one) to begin with as i would like to get that right before i go on to the directions (bearings) which if calculated from wrong distance information are just wasted time and effort, something i feel frustrated about at present.
so from part one my distance figures.
these of course are calculated as per the right angled triangle method as shown in part one.
point distance walked altogether distance in a straight line from O
O 0 0
A 1 1
B 3 sq rt of 5
C 6 (shown as 4 in part two) i have used 6) sq rt of 8
D 10 sq rt of 8
E 15 " " " 13 (this one agrees with an example given in part one)
F 21 " " " 25
G 28 " " " 32
H 36 " " " 32
I 45 " " " 41
J 55 " " " 61
though i have gone further than this i would like to get this (part one) correct in my mind before proceeding further again.
hope you understand my reluctance to spend too much time without some degree of assurance that i am on the right track (or not as the case may be.)
sincerely
David (diamond777)
I have worked hard on both of the above and having completed "1" - the distance one and being well advanced with "2" - the angle/bearing one ..........
would someone please tell me if i am on the right track.
May i point out that there appears to be a difference between the figure shown for distance walked altogether up to point C
in part "1" this is shown as 6 (kilometers)
whereas in part "2" the same figure is shown as 4 (kilometers)
I have proceeded using the figure from part "1" - namely 6 kms.
My calculated figures for the three figure bearings for points C, D, and E are respectively 348.7, 264.3 and 265.9 (all degrees of course)
Please advise if these figures of mine are correct. If they are I will proceed to attempt to calculate the similar figures for points F G H I and J which are calculated using polar coordinates rather than the cartesian coordinates used for points A to E.
My thanks in anticipation...........
diamond777
many thanks bob bundy
your reply should enable me now to get lots of practice........
i have completed problems in "triangle solving practice" and need to know how to access the answers to check if i have solved them correctly.
how do i access the answers please?
how do i find the answer/s to check if i have solved the problem/s correctly please? (see subject line - "triangle solving practice" in geometery section of maths is fun.)
why is the sq rt of 243 "9 sq rt of 3"? why not just nine?
seventy years young student of maths is fun
In need of lots of help!
lets start with:
algebra - working with exponents and logarithms-
question five
solve (for x)
2log3(x)=5
answer D x = 9square root(3)
i presume the 2 means that x is squared in the question?
please explain what 9square root of (3) is and how it is the square root of 243?
thank you ....
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