Trig geometry question (1 Viewer)

[YBK]

Hey, I can't do this question, any help would be appreciated.


Here's the question:


PQR is an isosceles triangle inscribed in a circle with centre O of radius one unit.

Let angle QOR = 2x, where x is an acute angle

a) show that the area A of Triangle PQR is given by A=sinx(cosx + 1)

b) Hence show that, as x varies, triangle PQR has its maximum possible area when it is equilateral.



I thought of a few ideas, but I think my method (if it works) will take too much time and is not the correct one to use. I split the circle up into three triangles, bah, confusing.



Thanks

lol, it refuses to show what x is

Edit: figured it out apparently u can't use that < sign

 

[Riviet]

Let... x be what?

 

[YBK]

Let... x be what?

tan x/2 ?

omg, can it be the t-formulas...
arggg, i'm soo lost..

 

[Riviet]

You haven't specified which angle x is...

 

[YBK]

You haven't specified which angle x is...

holy crap, ur right....

I'll edit the post, stupid me.

 

[pLuvia]

lol, wouldn't you realise from the original question

I was also confused

 

[YBK]

lol, wouldn't you realise from the original question

I was also confused

lol, I actually did write it, but it refused to show up... just check how many times I tried editing the post..

< ABC (try typing that in a post without a space and it wont show up) weird lol

 

[Riviet]

Draw OS, such that OS is perpendicular to QR. Then QS=RS since OS bisects QR.
Angle QOS = angle ROS = x, since OS bisects angle QOR. In triangle SOR, sinx=SR/OR
=SR, since OR is the radius of circle.
cosx=OS/OR
=OS
Now OS+PO= PS, which is the perpendicular height of triangle PQR.
But PO is the radius of the circle, which equals 1
so PS=OS+1
=cosx+1
Area of triangle PQR = 1/2.base.height
=SR.PS
=sinx(cosx+1) QED

 

[Riviet]

A=sinx.cosx+sinx
dA/dx=sinx.(-sinx)+cos²x+cosx
Area is a max when dA/dx=0,
ie cos²x-1+cos²x+cosx=0
2cos²x+cosx-1=0
(2cosx-1)(cosx+1)=0
Noting that 0< x <90^o,
cosx=1/2
x=60^o, which is angle QPR, which makes triangle PQR equilateral. Therefore area of triangle PQR is a max when x=60^o. -QED

 

[YBK]

Draw OS, such that OS is perpendicular to QR. Then QS=RS since OS bisects QR.
Angle QOS = angle ROS = x, since OS bisects angle QOR. In triangle SOR, sinx=SR/OR
=SR, since OR is the radius of circle.
cosx=OS/OR
=OS
Now OS+PO= PS, which is the perpendicular height of triangle PQR.
But PO is the radius of the circle, which equals 1
so PS=OS+1
=cosx+1
Area of triangle PQR = 1/2.base.height
=SR.PS
=sinx(cosx+1) QED

w00t! Go Riv!! Thanks

 

[YBK]

A=sinx.cosx+sinx
dA/dx=sinx.(-sinx)+cos²x+cosx
Area is a max when dA/dx=0,
ie cos²x-1+cos²x+cosx=0
2cos²x+cosx-1=0
(2cosx-1)(cosx+1)=0
Noting that 0< x <90^o,
cosx=1/2
x=60^o, which is angle QPR, which makes triangle PQR equilateral. Therefore area of triangle PQR is a max when x=60^o. -QED

makes sense , but I'm pretty sure you have to prove to the marker that u know that area could not be the minimum when dA/dx = 0

what I mean is that you have to prove that dA/dx = 0 is the max area. I'm sure you know how to do that, but it's just that my teacher really emphasises on it. I just draw a number line, and put the points where dA/dx = 0. Then I test a value to the left of the point and to the right of it, if to the right is negative, and to the left is positive, therefore it's a maximum turning point, which corresponds to the max area.

 

[pLuvia]

Yeh I know what you mean, but I'm sure Riviet would actually check it is a maximum in real life situation, you don't really need to show that it is a maximum, just find the f''(x) then say at that point f''(x)<0 therefore maximum

 

[YBK]

Yeh I know what you mean, but I'm sure Riviet would actually check it is a maximum in real life situation, you don't really need to show that it is a maximum, just find the f''(x) then say at that point f''(x)<0 therefore maximum

Yup, I'm sure Riviet would do it too

and yeah, you can do it through the second derrivative.