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.
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.
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
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!]/
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!]/
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
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
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
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
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
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