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can you explain the process o verification please
Aap explain bohat acha kartay hain boii!
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can you explain the process o verification please
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 youAap explain bohat acha kartay hain boii!
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
yeahThanks a lot for your help.. but the answer to the question is 74.4 and 338.7
Oh thats not a problem, dnt worryyeah
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
hahaha.... i am here to agree to that syed1995i 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!
Naah don't be sorryOh 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
Do yourself a favour. Watch these three videos: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 ??
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 ??
Thanks a lot for the great explanation. Appreciate it..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!
ThanksDo 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.
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