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Help with GCSE Maths question

Does any1 know how 2 do this
Reply 1
Please attach the question
Reply 2
some odd glitch i cant upload it but if u want to help then its Q19 and Q22 on: https://corbettmaths.com/wp-content/uploads/2019/04/Higher-Set-A-Paper-1.pdf
Original post by Gw.arz
some odd glitch i cant upload it but if u want to help then its Q19 and Q22 on: https://corbettmaths.com/wp-content/uploads/2019/04/Higher-Set-A-Paper-1.pdf


For 22) think about using the cos rule (twice) by drawing the line BD.
Reply 4
For 19, first work out the vector BD then work out FD as u now know BD. Then write GD as a multiple of FD eg lambda FD. Then write GC as a multiple of GE eg GC = xB. Then, work out 2 different routes which get you GD one of which uses GC and solve a simultaneous equation to work out lamdba and hence GC
Reply 5
For 22 work out BD in terms of x using the cosine rule. Then do cosine rule again to work out BAD now you have BD.
For those saying to use the cosine rule to find BD - there is a much quicker way to figure out the length of line BD.
Original post by alifelonglearner
For those saying to use the cosine rule to find BD - there is a much quicker way to figure out the length of line BD.


You don't really need to find BD, just set up the equation so
2*5^2 - 2*25^2cos(A) = 2x^2 - 2x^2/2
which is pretty much the answer. You can bisect BD and use right triangles as its a kite so 2 isosceles triangles as I guess youre suggesting, but can't see its much quicker than 1 or 2 lines + a small rearrangement?
(edited 1 year ago)
Original post by Gw.arz
some odd glitch i cant upload it but if u want to help then its Q19 and Q22 on: https://corbettmaths.com/wp-content/uploads/2019/04/Higher-Set-A-Paper-1.pdf


You need to post what you have tried - it's against forum rules for posters to do questions for you.
Original post by mqb2766
You don't really need to find BD, just set up the equation so
2*5^2 - 2*25^2cos(A) = 2x^2 - 2x^2/2
which is pretty much the answer. You can bisect BD and use right triangles as its a kite so 2 isosceles triangles as I guess youre suggesting, but can't see its much quicker than 1 or 2 lines + a small rearrangement?


What kind of triangle is BCD? :smile:
Original post by alifelonglearner
What kind of triangle is BCD? :smile:


Equilateral. Its still little different between doing 2x^2 - x^2 = x^2 or doing a bit extra spotting.
Original post by mqb2766
Equilateral. Its still little different between doing 2x^2 - x^2 = x^2 or doing a bit extra spotting.

In the long-run, spotting simple facts to reduce, even trivially, the complexity of the algebra needed to solve a problem is always recommended!
Original post by alifelonglearner
In the long-run, spotting simple facts to reduce, even trivially, the complexity of the algebra needed to solve a problem is always recommended!

If the standard way is striaghtforward, why bother? The main "complexity" in the question was drawing BD and using that to equate the info in the two triangles. After that, its fairly routine whichever you do it.
Original post by mqb2766
If the standard way is striaghtforward, why bother? The main "complexity" in the question was drawing BD and using that to equate the info in the two triangles. After that, its fairly routine whichever you do it.


In my experience, too many people leap into setting up the algebra without taking an extra moment to simplify the problem, ending up with extra variables they don’t need and inadvertently making the algebra more complicated than the problem itself.

In this case, yes, the extra algebra is trivial, but my point is more about picking up good habits.
Original post by alifelonglearner
In my experience, too many people leap into setting up the algebra without taking an extra moment to simplify the problem, ending up with extra variables they don’t need and inadvertently making the algebra more complicated than the problem itself.

In this case, yes, the extra algebra is trivial, but my point is more about picking up good habits.

Your original point seemed to be that it was much quicker to solve the problem doing this. Using the cos rule (twice) and equating BD^2 is pretty much as quick/simple. For this question, the "extra" algebra is hardly complex.
Original post by mqb2766
Your original point seemed to be that it was much quicker to solve the problem doing this. Using the cos rule (twice) and equating BD^2 is pretty much as quick/simple. For this question, the "extra" algebra is hardly complex.


My original and current point is - always take the easiest and simplest route! :smile:
Reply 16
Guys i found the mark scheme but i don't understand it. Its Q19 and Q22
https://corbettmaths.com/wp-content/uploads/2019/04/Higher-A-1-Answers-2-1.pdf
Original post by Gw.arz
Guys i found the mark scheme but i don't understand it. Its Q19 and Q22
https://corbettmaths.com/wp-content/uploads/2019/04/Higher-A-1-Answers-2-1.pdf

Can you explain for each question at which point you don't understand what the mark scheme is doing?
For Q22, remember that using trigonometry to calculate angles requires triangles (that statement may be less true the further you get in maths, but it works for now).

Make a sketch of triangle ABD, showing clearly the length of each side and indicating the angle you wish to calculate. Have you done that?

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