A maths - R-formulae (Trigo)
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Find all the angles between 0deg and 360 deg which satisfy the following equations.
sin x + 2 cos x = sq.rt 2
I got x = -24.20 and 77.3 deg. However the answer writtten is 77.3 and 335.85 deg.
Please examine my steps.
R = sq.rt 5
tan alfa = 2
alfa = tan-1 (2)
= 63.434
.: sin x + 2 cos x = sqrt. 5 sin(x +63.434)
sin (x + 63.434) = sqrt 2 / sqrt 5
1. (x + 63.434) is in 1st / 2nd quad
following which the new range is 63.434 < x + 63.434 <423.4
manipulate with the terms and i get -24.20 and 77.3 deg.
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Hi bonkysleuth,
The question requires that find all the angles between 0deg and 360 deg.
So, when your answer x = -24.20 ie a negative angle needs to be converted into a positive angle ie 360 degrees - 24.2 degrees = 335.8 degrees, you get the other answer. Your answer is actually correct, you just need to do the conversion.
Regards,
ahm97sic
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sian... recently a bit too busy to update examworld with more questions kindly posted out, and also questions kindly answered by Ahm97sic :(
Will get to it when I hv more time. :D
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Pardon me, but what is sq.rt? And alfa is actually alpha right....
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sq rt means square root lah.
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Originally posted by Ahm97sic:
Hi bonkysleuth,
The question requires that find all the angles between 0deg and 360 deg.
So, when your answer x = -24.20 ie a negative angle needs to be converted into a positive angle ie 360 degrees - 24.2 degrees = 335.8 degrees, you get the other answer. Your answer is actually correct, you just need to do the conversion.
Regards,
ahm97sic
Yea, but I don't understand why we have to convert it to positive angles whatsoever since this question doesn't require to find 2 rounds, etc.
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Hi bonkysleuth,
The conversion of negative angle to positive angle has nothing to do with ASTC or second round ie between 360 degrees and 720 degrees.
The question requires us to find all the angles between 0 degree and 360 degrees.
Since one of the answer is - 24.2 degrees, a negative angle, there is the need to convert it into a positive angle 360 - 24.2 = 335.8 degrees and the angle 335.8 degrees is between 0 degree and 360 degrees. So, the other answer is 335.8 degrees.
Regards,
ahm97sic
PS : bonkysleuth, it seemed that you have not practised trigonometry questions on negative angles. There are questions that asked students to find all the angles between - 180 degrees and 180 degrees. Please refer to Shing Lee Add Maths Textbook Page 173 Exercise 6D Question 6(a) to 6(f) for more practices.
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Originally posted by bonkysleuth:
Yea, but I don't understand why we have to convert it to positive angles whatsoever since this question doesn't require to find 2 rounds, etc.
OK you run a modelling agency and your client tell you "I want models between 170cm and 180cm to model my 10cm high heel shoes."
You then put a model on the 10cm shoes and measured her... ok 174cm. Satisfied the requirement of 170 to 180cm tio bo?
Then your client kpkb at you... because the range of height WITHOUT shoes is 170 to 180. The range of height WITH shoes is ACTUALLY 180 to 190cm.
You got sqrt. 5 sin(x +63.434) = sqrt. 2.
THE SUBJECT IS (x+63.434) NOT x at the point you looking at the SATC diagram.
Since 0<x<360
Then 63.434<x+63.434<423.434
Thus you SHOULD NOT have accepted the 1st solution AS it is not between 63.434 and 423.434. And you should have accepted the 3rd solution.
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Originally posted by Mikethm:
OK you run a modelling agency and your client tell you "I want models between 170cm and 180cm to model my 10cm high heel shoes."
You then put a model on the 10cm shoes and measured her... ok 174cm. Satisfied the requirement of 170 to 180cm tio bo?
Then your client kpkb at you... because the range of height WITHOUT shoes is 170 to 180. The range of height WITH shoes is ACTUALLY 180 to 190cm.
You got sqrt. 5 sin(x +63.434) = sqrt. 2.
THE SUBJECT IS (x+63.434) NOT x at the point you looking at the SATC diagram.
Since 0<x<360
Then 63.434<x+63.434<423.434
Thus you SHOULD NOT have accepted the 1st solution AS it is not between 63.434 and 423.434. And you should have accepted the 3rd solution.
TS, you understand ? I blur leh.
The question says find all angles between 0 and 360 but he says answer must be 423.434. So, question wrong huh ? answer provided by your teacher or book wrong huh ? or his answer wrong huh ?
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Originally posted by Lee012lee:
TS, you understand ? I blur leh.
The question says find all angles between 0 and 360 but he says answer must be 423.434. So, question wrong huh ? answer provided by your teacher or book wrong huh ? or his answer wrong huh ?
If you taking the 2008 exams, good luck to you.
The range in the question say to find all angles betweeen 0 and 360 HOWEVER it is referring to the value of "x".
When you are considering the SATC diagram, you are looking for the value of "x+63.434".
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Originally posted by Mikethm:
If you taking the 2008 exams, good luck to you.
The range in the question say to find all angles betweeen 0 and 360 HOWEVER it is referring to the value of "x".
When you are considering the SATC diagram, you are looking for the value of "x+63.434".
Hi Mikethm
Taken from bonkysleuth's answer
sin (x + 63.434) = 0.632456 (from sin (x + 63.434) = sqrt 2 / sqrt 5)
Basic angle (or Reference angle), theta
= sin-1 (0.632456)
= 39.232 degrees
Since sin +, so it lies in the 1st and 2nd quadrant,
x + 63.434 = 39.232, 180 - 39.232
x + 63.434 = 39.232, 140.768
x = 39.232 - 63.434, 140.768 - 63.434
x = - 24.2 , 77.33
Since x = - 24.2 degrees, it is a negative angle, we will need to convert it into a
positive angle 360 - 24.2 = 335.8 degrees.
Since x = 77.33 degrees and x = 335.8 degrees are between 0 degree and 360,
degrees. So, the answers are x = 77.33 degrees and x = 335.8 degrees
Check Answer :
(a) Substitute x = 77.33 degrees into
sin (x + 63.434) = sin (77.33 + 63.434) = 0.632456 (Correct)
(b) Substitute x = 335.8 degrees into
sin (x + 63.434) = sin (335.8 + 63.434) = 0.632456 (Correct)
(c) Using Mikethm's answer, substitute x = 423.434 into
sin (x + 63.434) = sin (423.434 + 63.434) = 0.8 (Incorrect)
(Hi Mikethm, you probably have made a mistake in your answer.)
Thank you for your kind attention.
Regards,
ahm97sic
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Originally posted by Ahm97sic:
Hi Mikethm
Taken from bonkysleuth's answer
sin (x + 63.434) = 0.632456 (from sin (x + 63.434) = sqrt 2 / sqrt 5)
Basic angle (or Reference angle), theta
= sin-1 (0.632456)
= 39.232 degrees
Since sin +, so it lies in the 1st and 2nd quadrant,
x + 63.434 = 39.232, 180 - 39.232
x + 63.434 = 39.232, 140.768
x = 39.232 - 63.434, 140.768 - 63.434
x = - 24.2 , 77.33
Since x = - 24.2 degrees, it is a negative angle, we will need to convert it into a
positive angle 360 - 24.2 = 335.8 degrees.
Since x = 77.33 degrees and x = 335.8 degrees are between 0 degree and 360,
degrees. So, the answers are x = 77.33 degrees and x = 335.8 degrees
Check Answer :
(a) Substitute x = 77.33 degrees into
sin (x + 63.434) = sin (77.33 + 63.434) = 0.632456 (Correct)
(b) Substitute x = 335.8 degrees into
sin (x + 63.434) = sin (335.8 + 63.434) = 0.632456 (Correct)
(c) Using Mikethm's answer, substitute x = 423.434 into
sin (x + 63.434) = sin (423.434 + 63.434) = 0.8 (Incorrect)
(Hi Mikethm, you probably have made a mistake in your answer.)
Thank you for your kind attention.
Regards,
ahm97sic
No I am not wrong.
sin x + 2 cos x = sq.rt 2
sin x + 2 cos x = sqrt. 5 sin(x +63.434)
sqrt. 5 sin(x +63.434) = sq.rt 2
sin(x +63.434) = sq.rt (2/5)
Basic angle = sin-1 [sq.rt (2/5)]
= 39.232
Solution is in 1st or 2nd quadrant since sin is +ve
Since 63.434 < x + 63.434 <423.4
We will reject the 1st solution of 39.232 (not in range), accept 140.768 and the 3rd solution of (360+39.232 aka 399.232)
Thus (x+63.434) = 140.768 or 399.232
x = 77. 334 or 335.798
x= 77.3 or 335.8
In addition, we do NOT ever check solutions by using equations we form (mistakes might already have been made along the way). Always use the original equation which is sin x + 2 cos x = sq.rt 2.
I have a >90% distinction rate with students who have been tutored by me for more than a year before their O levels. A 100% B3 or better for those tutored for >0.5 years. This is based on a brain dead list of standard approaches for 30 questions which will definitely cover 80% of the allocated marks... and the R formula is one of them (prior to 2002 and happily again from this year on). Those longer than a year manage distinctions because if we have time, then I encourage them to think why why why which better their chance of answering the thinking questions (20%)...
By the way, practically all my students failed their A Maths before I tutor them. I like to tutor students who failed because it is fun and challenging.
Oh did I mention no one ever failed on me in more than a decade of tutoring despite me being stubborn and not ever giving up on any student at all (except those who want to turn up less than once a week).
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Originally posted by Mikethm:
I have a >90% distinction rate with students who have been tutored by me for more than a year before their O levels. A 100% B3 or better for those tutored for >0.5 years. This is based on a brain dead list of standard approaches for 30 questions which will definitely cover 80% of the allocated marks... and the R formula is one of them (prior to 2002 and happily again from this year on). Those longer than a year manage distinctions because if we have time, then I encourage them to think why why why which better their chance of answering the thinking questions (20%)...
By the way, practically all my students failed their A Maths before I tutor them. I like to tutor students who failed because it is fun and challenging.
Oh did I mention no one ever failed on me in more than a decade of tutoring despite me being stubborn and not ever giving up on any student at all (except those who want to turn up less than once a week).
So have my students for both maths and physics
Right now, I have a challenge of someone who started tutoring with me only 3 lessons before prelims :( (from failing)
I agree that brain dead training is useful for desperate situtations, but if you have >0.5 years with them, that shouldn't be the way. This method might kill them in A levels if they go JC, or if it doesn't kill them at A levels, it will kill them at O levels. Many a times, I have seen students who struggled at A levels because their foundation at O levels wasn't well set; O levels is just too easy to score (for maths and physics). And when they reach uni levels, the system will wipe them out if they continued this method.
That's the reason why I like UltimaOnline's way of helping in this forum; giving lots of extra information on top of explaining/guiding students to think.
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Originally posted by eagle:
So have my students for both maths and physics
Right now, I have a challenge of someone who started tutoring with me only 3 lessons before prelims :( (from failing)
I agree that brain dead training is useful for desperate situtations, but if you have >0.5 years with them, that shouldn't be the way. This method might kill them in A levels if they go JC, or if it doesn't kill them at A levels, it will kill them at O levels. Many a times, I have seen students who struggled at A levels because their foundation at O levels wasn't well set; O levels is just too easy to score (for maths and physics). And when they reach uni levels, the system will wipe them out if they continued this method.
That's the reason why I like UltimaOnline's way of helping in this forum; giving lots of extra information on top of explaining/guiding students to think.Well you gotta understand why I said what I said then. Those R formula questions are so "dead" that it is insulting to imply that anyone can get them wrong.
And even if I have <0.5 years with them, I will not resort to brain dead training solely. I always explain the concepts (hastily) even if we are damned desperate.
Tuition is a game to me. The game of trying to get 100% distinctions for a year. Which I had failed to so far... best record was 10 distinctions outta 11... sob sob. Thus I will never resort to 100% brain dead training. Anyone who is brain dead will NEVER get a distinction. Which is why I always tell my students... any distinction you get will be because of you and not because of me.
However, if you think throughout the entire 2.5hrs exams... you are looking for trouble as well as you will be mentally exhausted near the end. Which is why I insist on don't think if you don't need to, think only when necessary.
Today's schedule.
9am-2pm. Getting students to explain the concepts to me PLUS guide me thru the 30 sequences
3pm-9pm. Modified 2006 Paper I & II
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Originally posted by Mikethm:
Well you gotta understand why I said what I said then. Those R formula questions are so "dead" that it is insulting to imply that anyone can get them wrong.
btw, I saw your google ad on your website on my ExamWorld
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Originally posted by eagle:
btw, I saw your google ad on your website on my ExamWorld
Ya, I signed up with a Netfirm hosting plan (backup hosting plan) for $10 and it came with a $50 google ad coupon... tried it out to see how much Google made. What the hell... your bloody click cost me $0.80. And they paid me $0.01 for clicks on my website... now I know why Google is a multibillion company.
Anyway, if you want el cheapo hosting for your examworld plus domain registration, the netfirm $10 per year plan come with 2 free domain names. you can get it at http://www.netfirms.com/mac
Sometimes those domain and hosting companies defy mathematical logic... Godaddy want to charge me $200+ for registering 2 domain names for 7 years with privacy protection.
Then I noticed they offer free privacy protection for registering 5 domains or more. And then I notice they are offering .info domains for $0.99 for 1st year.
So I ended up renewing 2 domains with 7 years' free privacy protection + 3 .info domains for $100... knn if the E Maths paper come out Godaddy example of real life application, all die cock stand sia...
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Originally posted by Mikethm:
Ya, I signed up with a Netfirm hosting plan (backup hosting plan) for $10 and it came with a $50 google ad coupon... tried it out to see how much Google made. What the hell... your bloody click cost me $0.80. And they paid me $0.01 for clicks on my website... now I know why Google is a multibillion company.
Anyway, if you want el cheapo hosting for your examworld plus domain registration, the netfirm $10 per year plan come with 2 free domain names. you can get it at http://www.netfirms.com/mac
Sometimes those domain and hosting companies defy mathematical logic... Godaddy want to charge me $200+ for registering 2 domain names for 7 years with privacy protection.
Then I noticed they offer free privacy protection for registering 5 domains or more. And then I notice they are offering .info domains for $0.99 for 1st year.
So I ended up renewing 2 domains with 7 years' free privacy protection + 3 .info domains for $100... knn if the E Maths paper come out Godaddy example of real life application, all die cock stand sia...
when u click?
I would want hosting, but probably in the future... Most likely I will continue to use blogger as the host, but get a domain to point to blogger. It's a good and free CMS for me :D