How do i solve this maths problem???????? Someone please!
Can someone solve the following question that is in this picture???????? My maths teacher couldnt solve it!!
Posted from TSR Mobile
Posted from TSR Mobile
How about you try working out the length of AB. M being the midpoint should be a big clue.
I dont think its possible knowing just the 2 angles and not one of the sides
Also: we are not allowed to use a calculator!!
Posted from TSR Mobile
Also: we are not allowed to use a calculator!!
Posted from TSR Mobile
You need to subtend a perpendicular line from C to produce two right angle triangles with A and B, call the point where the perpendicular line meets AB point D.
Now using the fact that tan x = opp/adj you can find the length of CD.
tan B = 2 / 3
tan A = 1 / 6 = 2 / 12
So since the line CD is similar to both right angle triangles we can see it is 2.
Now the base AB is clearly 3 + 12 = 15.
Therefore AM is 15 / 2 = 7.5
So, MD is 12 - 7.5 = 4.5
Using the tan x = opp / adj again:
tan (theta) = 2 / 4.5 = 4 / 9
Now using the fact that tan x = opp/adj you can find the length of CD.
tan B = 2 / 3
tan A = 1 / 6 = 2 / 12
So since the line CD is similar to both right angle triangles we can see it is 2.
Now the base AB is clearly 3 + 12 = 15.
Therefore AM is 15 / 2 = 7.5
So, MD is 12 - 7.5 = 4.5
Using the tan x = opp / adj again:
tan (theta) = 2 / 4.5 = 4 / 9
Original post by HelloAdamLewis
You need to subtend a perpendicular line from C to produce two right angle triangles with A and B, call the point where the perpendicular line meets AB point D.
Now using the fact that tan x = opp/adj you can find the length of CD.
tan B = 2 / 3
tan A = 1 / 6 = 2 / 12
So since the line CD is similar to both right angle triangles we can see it is 2.
Now the base AB is clearly 3 + 12 = 15.
Therefore AM is 15 / 2 = 7.5
So, MD is 12 - 7.5 = 4.5
Using the tan x = opp / adj again:
tan (theta) = 2 / 4.5 = 4 / 9
Now using the fact that tan x = opp/adj you can find the length of CD.
tan B = 2 / 3
tan A = 1 / 6 = 2 / 12
So since the line CD is similar to both right angle triangles we can see it is 2.
Now the base AB is clearly 3 + 12 = 15.
Therefore AM is 15 / 2 = 7.5
So, MD is 12 - 7.5 = 4.5
Using the tan x = opp / adj again:
tan (theta) = 2 / 4.5 = 4 / 9
As much as I'm sure the OP appreciates your help, we are not supposed to just give out the answer. You are supposed to just lead them to it on here.
Just telling the OP to draw a perpendicular line would have been the next thing to do.
I'm by no means saying you are wrong - that's exactly how I did it. Just be aware that on TSR we aren't supposed to do the homework for people. Just help them by leading them towards it.
(edited 10 years ago)
Original post by Occams Chainsaw
As much as I'm sure the OP appreciates your help, we are not supposed to just give out the answer. You are supposed to just lead them to it on here.
Just telling the OP to draw a perpendicular line would have been the next thing to do.
I'm by no means saying you are wrong - that's exactly how I did it. Just be aware that on TSR we aren't supposed to do the homework for people. Just help them.
Just telling the OP to draw a perpendicular line would have been the next thing to do.
I'm by no means saying you are wrong - that's exactly how I did it. Just be aware that on TSR we aren't supposed to do the homework for people. Just help them.
You're not 'supposed' to do anything. He asked for an answer, the guy gave him one
Original post by pirateship
You're not 'supposed' to do anything. He asked for an answer, the guy gave him one
http://www.thestudentroom.co.uk/wiki/Study_Help_Guidelines
"Wherever possible try to guide the asker to the answer rather than just giving the answer straight off! Not much is learned by doing this and the asker may not understand why the answer is what it is.
This Socratic Method of learning - where people ask a question and learn the answer to it by directed prodding and questioning - leads to the best results and understanding.
Not all types of question lends itself to this type of technique but where possible it should be employed and full solutions (especially to calculations) should only be posted as a last resort or a significant misunderstanding has developed. It is at the moderators' discretion to remove, edit or even warn posts that flout this rule, particularly if the post occurs after someone has tried to guide them through the answer."
I know u probably mean well but this isnt a homework question...
Posted from TSR Mobile
Posted from TSR Mobile
Even more reason for us to not just give out the answer and to help you get there yourself.
Also, if I wanted homework help and people weren't supposed to give it I'm sure I'd say it isn't homework.
Anyway, I hope you understand how it was done. If not I, or someone else, can go through it with you now the cat is out of the bag.
Also, if I wanted homework help and people weren't supposed to give it I'm sure I'd say it isn't homework.
Anyway, I hope you understand how it was done. If not I, or someone else, can go through it with you now the cat is out of the bag.
Original post by afnanm1
Can someone solve the following question that is in this picture???????? My maths teacher couldnt solve it!!
Posted from TSR Mobile
Posted from TSR Mobile
AHHH this question from the BMAT last year! I sat this exam, it was the TOUGHEST EXAM EVER
I was on this question for like 3 minutes and just gave up. Nobody from my school could do it either....Original post by Occams Chainsaw
As much as I'm sure the OP appreciates your help, we are not supposed to just give out the answer. You are supposed to just lead them to it on here.
Just telling the OP to draw a perpendicular line would have been the next thing to do.
I'm by no means saying you are wrong - that's exactly how I did it. Just be aware that on TSR we aren't supposed to do the homework for people. Just help them by leading them towards it.
Just telling the OP to draw a perpendicular line would have been the next thing to do.
I'm by no means saying you are wrong - that's exactly how I did it. Just be aware that on TSR we aren't supposed to do the homework for people. Just help them by leading them towards it.
I'm very sorry if I've deprived the OP of a chance to discover the solution for himself. I was just replying to his specific request for a solution. If someone were to ask for help, I would always avoid ruining the solution for them.
I appreciate your respect for the rules.
Original post by HelloAdamLewis
I'm very sorry if I've deprived the OP of a chance to discover the solution for himself. I was just replying to his specific request for a solution. If someone were to ask for help, I would always avoid ruining the solution for them.
I appreciate your respect for the rules.
I appreciate your respect for the rules.
I don't care about the rules but I do care about people being able to understand how the maths is done!
I just use that rule to stop people giving out the answer straight away. Its no problem, really. I just know that I learn much better by working as much of a question out myself as possible.
Original post by Occams Chainsaw
I don't care about the rules but I do care about people being able to understand how the maths is done!
I just use that rule to stop people giving out the answer straight away. Its no problem, really. I just know that I learn much better by working as much of a question out myself as possible.
I just use that rule to stop people giving out the answer straight away. Its no problem, really. I just know that I learn much better by working as much of a question out myself as possible.
I completely agree
As a note to my solution:
The lengths that I used are not the only possible lengths; in fact any constant multiplied by the side lengths can be used because the trig. functions give a ratio of the sides.
This is why I changed one of the fractions to 2 / 12.
My explanation isn't very clear, but it's quite hard to describe.
Original post by HelloAdamLewis
I completely agree
As a note to my solution:
The lengths that I used are not the only possible lengths; in fact any constant multiplied by the side lengths can be used because the trig. functions give a ratio of the sides.
This is why I changed one of the fractions to 2 / 12.
My explanation isn't very clear, but it's quite hard to describe.
As a note to my solution:
The lengths that I used are not the only possible lengths; in fact any constant multiplied by the side lengths can be used because the trig. functions give a ratio of the sides.
This is why I changed one of the fractions to 2 / 12.
My explanation isn't very clear, but it's quite hard to describe.
No. that makes perfect sense (to me). The fact that the answers were all tan is what gave it away to me. You were meant to realise that all you needed was the ratio.
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