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Where's 'x'??
Don't you think you missed something while posting this...?
Nice point!
However, 0`=360`=2n*pi
But i think 0` is Acute..
You mean that E is the point of intersection of AD & BC"extended"?
Or did you mis-spell AC & BD for AD & BC?
Because AC & BD form the chords of the circle and intersect too but for chords AD & BC to intersect, they need to be extended to meet at some point outside the circle...
Another way to do it would be..
Let..
rt(2 + rt3) = x + rty
2 + rt3 = x^2 + y + 2x(rty)
x^2 + y = 2 ...(1)
&
2x(rty) = rt3
---> y = 3/(4x^2) ...(2)
putting in (1)
x^2 + 3/(4x^2) = 2 which is quadratic in x^2.
Considering only positive roots, we get..
x = rt(1/2) & rt(3/2)
which gives
y = rt(3/2) & rt(1/2)
etc. etc.
Thus simplify all the surds & you'll get the answer!!!
Regards...
Consider
2 + rt3 = 1/2 + 3/2 + 2 1/2 rt3
= [rt(1/2)]^2 + [rt(3/2]^2 + 2 rt(3/2) rt(1/2)
---> 2 + rt3 = [rt(1/2) + rt(3/2)]^2
therefore..
rt(2 + rt3) = rt(1/2) + rt(3/2)
similarly..
rt(4 - rt7) = rt(7/2) - rt(1/2)
rt(5 + rt21) = rt(7/2) + rt(3/2)
substitute the values in LHS & RHS n its proved!!
Kindly don't mind my not so neat method of writing but actually i'm using my mobile to surf n don't really know how to write codes..
Got it Jane...
cos6x = 0.5 = cos60 = cos(360+60) = cos(360-60)
cos6x = cos60 = cos420 = cos300
x = 10 , 70 , 50
right?!
Thanks!
You've got a bird's eye...
Hmmm...!!
One i've spotted!
From step 3, we can write..
sin5x sin4x - 2 sin3x sin4x cos4x = 0
sin4x (sin5x - 2 sin3x cos4x) = 0
so..
sin4x = 0 = sin 180
---> x = 45`
or
sin5x - 2sin3x cos4x = 0
---> which (as per my method) gives x = 60`
thinking about other two solutions!
Jane is too intelligent!
I shoulda been Tarzan!!
sin5x sin4x = sin3x sin8x
sin5x sin4x = sin3x 2sin4xcos4x
sin5x = 2 sin3x cos4x
sin5x = sin (3x+4x) + sin (3x - 4x)
sin5x = sin7x - sinx
sin7x - sin5x = sinx
sin (6x+x) - sin (6x-x) = sinx
2cos6xsinx = sinx
2cos6x = 1
cos6x = 0.5 = cos 60`
x = 10`
so.. x = 10 degrees!!
Was quite easy, right??
Wonder why some posts remain unanswered??
The answer to QUESTION 10 can be...
( 11 ! ) ^ ( 11 ! )
thats 11 factorial raised to 11 factorial!!
But it can be extended to any number of factorials that one wishes to...
Question is INCOMPLETE i must say!?
Let a & b be the required factors..
So we must have..
ab = 28 ...(1)
a + b = 11 ...(2)
squaring (2) we get..
a^2 + b^2 + 2ab = 121
a^2 + b^2 = 121 - 2ab
put ab from (1) to get..
a^2 + b^2 = 121 - 2*28 = 65
now subtract 2ab from both sides to get..
a^2 + b^2 - 2ab = 65 - 2ab
(a - b)^2 = 65 - 56 = 9
--> a - b = 3 ...(3)
now add (2) & (3)..
(a + b) + (a - b)= 11 + 3
2a = 14
a = 7
putting this a in (2)..
b = 11 - 7
b = 4
hence, required factors are 7 & 4 !!
:-)
Very well said mathsyperson!
I'm thinking again..
There's ONLY ONE such number & hence it should be uniquely determinable!!
Consider for example..
Find a number which is square of its last digit (digit in unit's place)?
Ans (quite obvious) is 25 & 36 and these can be determined using almost the same logic as above, for solving the equation!
The required number can only be 'one/two digit' number can also be shown..
An 'n digit' number 'a' would satisfy the above question only if "square of the sum of the digits of the GREATEST 'n digit' number EQUALS/EXCEEDS the number'!
Greatest 'one digit' number is 9..
9^2 = 81 > 9... Satisfies!
Greatest 'two digit' number is 99..
9+9 = 18
18^2 = 324 > 99... Satisfies!
Greatest 'three digit' number is 999..
9+9+9 = 27
27^2 = 729 < 999.. Doesn't Satisfy!!
& so on...
Comments are most welcome!!
Regards...
Let no. be 10x+y (x,y=0-9;x=!0).
Now..
10x + y = (x+y)^2
9x = (x+y)^2 - (x+y)
let's say x + y = z
so..
9x = z^2 - z = z(z-1)
consider rhs.. Its a product of two CONSECUTIVE numbers!
So, 9x should be multiple of two CONSECUTIVE numbers & x MUST be a ONE DIGIT number which leads to only consecutive number to 9 i.e. '8'!
So.. x = 8 is the only solution which in turn gives y = 1!
So the number is 81!
Hi guys!
Firstly, i'm new to the site & i'm really pleased to see so many people interested/engrossed in Mathematics!
Maths, no doubts, is most Beautiful, Puzzling & Fathomless of all things!
I think i've found a nice Mathematical way to crack the above problem but i'll find it a bit difficult to post it coz i don't know how to code for equations n i'm surfing through my Nokia6600 rather than computer..
Here it goes...
Seen it!
Very well done!
Thanks once again...
Make the 1st cigarette out of 5 butts.
After smoking it, you get 1 butt. Add 4 more butts to it to make the 2nd cigarette.
After smoking the second cigarette, add 4 more butts to make the 3rd cigarette.
So if a[sub]n[/sub] is the cumulative total number of butts required to make the nth cigarette,
Hence a[sub]30[/sub] = 121.
Really Impressive!!
I'd been thinking over it for a few weeks but couldn't get it!!
More posts to come soon..
Love u.. Maths Forum!!
Smokie can not do without smoking & he has learned to make cigarettes outta butts!!
If 1 cigarette leaves behind 1 butt & 5 butts are used to make 1 new cigarette then whats the MINIMUM number of BUTTS he needs to smoke 30 cigarettes?
Answer is 121 butts & i verified its true & does allow him to smoke 30 cigarettes but how does this number come?
?
Please help...
81 is one possible answer.
yes.. It OBVIOUSLY is the answer but how does one WORK IT OUT? Mathematically & Logically??
Find a number (mathematically & logically) which is square of the sum of its digits!
How many digits can the number have?
"Zero is the real Hero, that's y i'm ZHero!"
Main puzzle is at the bottom....
Q: If 1 cigarette leaves 1 butt & 5 butts make a new cigarette then how many cigarettes can be made out of 121 butts?
A: 30! Its quite simple as 120 butts will make 24 cigarettes that would leave us with 24 butts which will combine with 1 butt left to make 5 more cigarettes which in turn will yield 1 more cigarette making a total of 30!
Now.. If the question be..
Find the LEAST no. of butts required to make 30 cigarettes if 1 cigarette leaves back 1 butt & 5 butts make 1 new cigarette?!
How to solve this??
"Zero is the real Hero, that's y i'm ZHero!"