Mathematics: Post your doubts here!

[syed1995]

can you explain the process o verification please

Aap explain bohat acha kartay hain boii!

 

[abruzzi]

Could someone please help me with part (iii) of the question? Its kinda urgent
The answer to part(ii) is √20sin(theta + 63.43)
Thanks

 

[A star]

Aap explain bohat acha kartay hain boii!

i am out of form B| ask daredevil . stupid computing project took me 8 sleepless nights. btw where the hell were you . I thought we would get to the magic no (100) in the convo without you

 

[midha.ch]

Could someone please help me with part (iii) of the question? Its kinda urgent
The answer to part(ii) is √20sin(theta + 63.43)
Thanks

 

[abruzzi]

Thanks a lot for your help.. but the answer to the question is 74.4 and 338.7

 

[midha.ch]

Thanks a lot for your help.. but the answer to the question is 74.4 and 338.7

yeah
subtract 63.43 from both values
check I have written 137.9 and 402.1 is equal to (Theta+ 63.43) not Theta
so for theta you have to subtract 63.43 from both sides
the last line didn't come in the pic where I subtracted
It was my careless mistake when I took the snap

 

[abruzzi]

yeah
subtract 63.43 from both values
check I have written 137.9 and 402.1 is equal to (Theta+ 63.43) not Theta
so for theta you have to subtract 63.43 from both sides
the last line didn't come in the pic where I subtracted
It was my careless mistake when I took the snap

Oh thats not a problem, dnt worry
I didnt get one part of the solution you posted.
theta + 63.43 = (inv)sin(3/√20)
hence, theta + 63.43 = 42.13
theta = -21.3

How did you get your values?

Sorry for bothering.. I'm a li'l weak in trigonometry

 

[daredevil]

i am out of form B| ask daredevil . stupid computing project took me 8 sleepless nights. btw where the hell were you . I thought we would get to the magic no (100) in the convo without you

hahaha.... i am here to agree to that syed1995

and yeahhh where were u??! the century is not something u want to miss dear sir ... esp not in the T-20 season it's important!

 

[syed1995]

hahaha.... i am here to agree to that syed1995

and yeahhh where were u??! the century is not something u want to miss dear sir ... esp not in the T-20 season it's important!

i was waiting for you guys to reach 99 to make an entry and bowl you guys down.. but you guys slowed down

anyways on topic, answer that question DEAR!

 

[midha.ch]

Oh thats not a problem, dnt worry
I didnt get one part of the solution you posted.
theta + 63.43 = (inv)sin(3/√20)
hence, theta + 63.43 = 42.13
theta = -21.3

How did you get your values?

Sorry for bothering.. I'm a li'l weak in trigonometry

Naah don't be sorry
these are the few twists in trigonometry which I think is literally pointless!
you see what you wrote is logically right! but we have a range given for theta.
Now sine ratio is +ve in 1st and 2nd quadrant.
we have a range given for Theta but we have value for sin(theta + 63.43)
we can't simply take sin inverse and subtract 63.43 from the angle as on both sides our variable isn't the same.
I'll make it bit more simple
consider Theta+63.43= Y
and you have sinY = 3/√20
but your given range is for Theta not Y
so convert Theta to Y
I have shown in the picture
63.43 < Y < 423.43
Now back to the solution you suggested
yes you are right! sin inverse of 3/√20 does equal to 42.13
but that's out of the given range of Y
that's why 42.13 is not acceptable
quick tip- when solving trig sums where range is not 0<variable used for angle<360 it's better if you draw the range cycle and mark out possible quadrants your angle could lie in (here since my sine ratio is +ve I have marked 1st and 2nd quadrant)
and when writing an answer make sure you mention ALL THE POSSIBLE VALUES thus I got 2 values

guess I wrote too long but you said you were weak in trig so I tried to explain it as clearly possible!

 

[sitooon]

Solve the inequality I x I < I 5 + 2x I
I got two values to x as -5 and -5/3 and they are right ,
but how to make the inequalities finally ..

someone already solved it , but can anyone give alternative way ? of finding the range of values of x ??

 

[Suchal Riaz]

Solve the inequality I x I < I 5 + 2x I
I got two values to x as -5 and -5/3 and they are right ,
but how to make the inequalities finally ..

someone already solved it , but can anyone give alternative way ? of finding the range of values of x ??

Do yourself a favour. Watch these three videos:
http://examsolutions.net/maths-revi...functions/modulus/inequalities/tutorial-1.php
http://examsolutions.net/maths-revi...functions/modulus/inequalities/tutorial-2.php
http://examsolutions.net/maths-revi...functions/modulus/inequalities/tutorial-3.php
If there's some problem like youtube is blocked i know you will figure out a way to see these videos.

 

[midha.ch]

Solve the inequality I x I < I 5 + 2x I
I got two values to x as -5 and -5/3 and they are right ,
but how to make the inequalities finally ..

someone already solved it , but can anyone give alternative way ? of finding the range of values of x ??

At first I got it in the form of a^2 - b^2
then simplified it
about how I got the three possible ranges on the table,
if it was an equation instead of inequality then x = -5 and -5/3
if drawn on a number line there are 3 possibilities of the correct range
so draw a table
you get the signs for (x + 5) by checking in each range if a value of x is substituted what result you get
for example in the range x< -5 , take a value less than -5
If we take -6 then (-6 +5) = -1 which is negative (thus the minus sign in the table for x+5 in the x<-5 coloumn )
so (x+5) is -ve when x< -5
check with the rest range possibilities
for (3x + 5) do the same thing
(x+5) is multiplied to (3x+5)
so multiply the corresponding signs
example (x+5)(3x+5) for x<-5 = -ve multiplied by -ve = +ve
My initial inequality was (x+5)(3x+5) > 0
that means the product should be +ve
from the table this happens only when x<-5 or x>-5/3

 

[abruzzi]

Naah don't be sorry
this are the few twists in trigonometry which I think is literally pointless!
you see what you wrote is logically right! but we have a range given for theta.
Now sine ratio is +ve in 1st and 2nd quadrant.
we have a range given for Theta
but we have value for sin(theta + 63.43)
we can't simply take sin inverse and subtract 63.43 from the angle as on both sides our ariable isn't the same.
I'll make it bit more simple
consider Theta+63.43= Y
and you have sinY = 3/√20
but your given range is for Theta not Y
so convert Theta to Y
I have shown in the picture
63.43 < Y < 423.43
Now back to the solution you suggested
yes you are right! sin inverse of 3/√20 does equal to 42.13
but that's out of the given range of Y
that's why 42.13 is not acceptable
quick tip- when solving trig sums where range is not 0<variable used for angle<360 it's better if you draw the range cycle and mark out possible quadrants your angle could lie in (here since my sine ratio is +ve I have marked 1st and 2nd quadrant)
and when writing an answer make sure you mention ALL THE POSSIBLE VALUES thus I got 2 values

guess I wrote too long but you said you were weak in trig so I tried to explain it as clearly possible!

Thanks a lot for the great explanation. Appreciate it..

 

[Failure]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w03_qp_5.pdf
#5(ii)
the value of a is 36.9

 

[Suchal Riaz]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w03_qp_5.pdf
#5(ii)
the value of a is 36.9

https://app.box.com/s/d0z8lcziqvz7651zvv17

 

[sitooon]

Do yourself a favour. Watch these three videos:
http://examsolutions.net/maths-revi...functions/modulus/inequalities/tutorial-1.php
http://examsolutions.net/maths-revi...functions/modulus/inequalities/tutorial-2.php
http://examsolutions.net/maths-revi...functions/modulus/inequalities/tutorial-3.php
If there's some problem like youtube is blocked i know you will figure out a way to see these videos.

Thanks

 

[Failure]

https://app.box.com/s/d0z8lcziqvz7651zvv17

thanks

 

[ahmed abdulla]

Failure .. see if you can help

 

[yara]

Please somebody explain how it is solved with its steps...many thanks
O/n 2013 9709 question 6