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Confusing Triangle Question

Hi,

Attached is the question that caught me out 9iv (i.e. the highlighted) question was the one that I was stuck with. Please could someone help.

Thanks a lottt!!

PS - As you can see from my previous posts, I like to pay people back for their help
Reply 1
Differentiate it and let it equal to zero to find a minimum (should work).
Reply 2
If you have the area of the triangle (part 3) in terms of a variable, you just need to change the variable so your expression is as small as possible - as your variable is part of the denominator, you want to make it as large as possible.
Reply 3
Original post by Zygroth
If you have the area of the triangle (part 3) in terms of a variable, you just need to change the variable so your expression is as small as possible - as your variable is part of the denominator, you want to make it as large as possible.


Or this... It's more straight forward.
OP- for the last part, you do NOT need calculus. How big can sin(theta) be, if theta is a real number? How small can it be?

Once you've thought about that, look at the result you've proved in (iii).
Reply 5
Original post by theOldBean
OP- for the last part, you do NOT need calculus. How big can sin(theta) be, if theta is a real number? How small can it be?

Once you've thought about that, look at the result you've proved in (iii).




Original post by limetang
Or this... It's more straight forward.



Original post by Zygroth
If you have the area of the triangle (part 3) in terms of a variable, you just need to change the variable so your expression is as small as possible - as your variable is part of the denominator, you want to make it as large as possible.


Could you show me the stages of both your methods pls
Original post by freedomhunter
Could you show me the stages of both your methods pls


I am reluctant to just give you the answer, because you would benefit more from trying to work it out yourself, given the hints you have received.

So here's a fully worked example to a closely related question:

"Find the minimum value of 2/cos(y)"

For all y, -1<cos(y)<1, with equality if y=0 (since cos 0 = 1). Of course there are other values that work too - in your problem you probs need to pick the value for theta that is 'most likely to be the angle in a triangle.'

So if we want to minimise 2/cos(y), we need to maximise cos(y). Since the maximum of cos(y) is 1, the minimum value is 2/1, which is 2, when y=0. [Note that choosing -1 isn't allowed in your problem as the area of a triangle cannot be negative!]/
Reply 7
Original post by theOldBean
I am reluctant to just give you the answer, because you would benefit more from trying to work it out yourself, given the hints you have received.

So here's a fully worked example to a closely related question:

&quot;Find the minimum value of 2/cos(y)&quot;

For all y, -1&lt;cos(y)&lt;1, with equality if y=0 (since cos 0 = 1). Of course there are other values that work too - in your problem you probs need to pick the value for theta that is 'most likely to be the angle in a triangle.'

So if we want to minimise 2/cos(y), we need to maximise cos(y). Since the maximum of cos(y) is 1, the minimum value is 2/1, which is 2, when y=0. [Note that choosing -1 isn't allowed in your problem as the area of a triangle cannot be negative!]/


This is what Ive done so far:
in order to find the least value of 12/sin2p, sin2p has to be at its greatest value. But im not sure what the greatest value of sin2p is and im not sure how to get it. Once Ive got that I think ill be ok to do the rest
Original post by freedomhunter
This is what Ive done so far:
in order to find the least value of 12/sin2p, sin2p has to be at its greatest value. But im not sure what the greatest value of sin2p is and im not sure how to get it. Once Ive got that I think ill be ok to do the rest


What does the graph of y=sin(2x) look like?
Original post by freedomhunter
This is what Ive done so far:
in order to find the least value of 12/sin2p, sin2p has to be at its greatest value. But im not sure what the greatest value of sin2p is and im not sure how to get it. Once Ive got that I think ill be ok to do the rest


When is sin2x at its maximum?
Reply 10
Original post by theOldBean
What does the graph of y=sin(2x) look like?


The highest point would be at 2? or am confusing it with 2sinx?

Edit: Ahh bummer, yes I am confusing it with 2sinx! Highest value of sin(2x) would be 1 and therefore least possible value is 12/1 = 12? Please tell me I am right otherwise Im gna feel like sucha fool :tongue:
(edited 12 years ago)
Original post by freedomhunter
The highest point would be at 2? or am confusing it with 2sinx?

Edit: Ahh bummer, yes I am confusing it with 2sinx! Highest value of sin(2x) would be 1 and therefore least possible value is 12/1 = 12? Please tell me I am right otherwise Im gna feel like sucha fool :tongue:


CORRECT! (well your edit is anyway)
and of course the value of p is such that sin(2p)=1, so 2p=pi/2, giving p=pi/4.

Edit: It's a nice feeling when you finally crack a problem, isn't it :wink:
Reply 12
LOOOOOL! Finally! haha thanksss :P

Reps for EVERYONE :biggrin:
Reply 13
Original post by theOldBean
CORRECT! (well your edit is anyway)
and of course the value of p is such that sin(2p)=1, so 2p=pi/2, giving p=pi/4.

Edit: It's a nice feeling when you finally crack a problem, isn't it :wink:


sorry - cant rep all your posts cos ive repped too many of your posts already :tongue:

Ive got a few more questions if you can help me crack those too (esp the parametric ones).

Im gna go sleep now cos im running out of brain juice so Im gna return to these questions that Ive posted tomorrow morning.

Please try the others questions :biggrin:

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