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Mathematics 8 Online
OpenStudy (callisto):

Trigo problem #2

OpenStudy (callisto):

OpenStudy (experimentx):

pi

OpenStudy (anonymous):

well first thing you should do is find x = pi/2

OpenStudy (anonymous):

to find the height of the triangle, then you need to find the x intercepts to find the length of the base

OpenStudy (callisto):

why is x=pi/2? I found it to be pi/6 and 11pi/6 ...

OpenStudy (callisto):

I meant for the x-ints

OpenStudy (anonymous):

then you can just use the standard formula bh/2 to find the area

OpenStudy (anonymous):

i found none of these

OpenStudy (anonymous):

i found \(A=\frac{1}{2}\)

OpenStudy (experimentx):

Well I found pi

OpenStudy (anonymous):

oh i mean i did not find these for A and B and C

OpenStudy (anonymous):

you would take x = pi/2, because that will give you the height of your triangle because it is at the triangles tip

OpenStudy (experimentx):

everybody is getting different answers. lol

OpenStudy (callisto):

So, the answer is C, if it is correct

OpenStudy (anonymous):

i did not find the answer yes you need A and B and C i got \(A=\frac{1}{2}\), \(C=3\), \(B=\pi-\frac{1}{2}\)

OpenStudy (kinggeorge):

First you find the height of the triangle to be \(y=1\). Now you need to find the base. Since \[0=1-2\cos(2x)\]At \(x=\pi/6\), we know the base is \(2\cdot (\pi/2 -\pi/6)\) = \(2\pi/3\)

OpenStudy (kinggeorge):

To find the area of the triangle, we multiply by height, which is 1, and divide by 2 to get \(\pi/3\).

OpenStudy (anonymous):

i am way way wrong nvm

OpenStudy (callisto):

Oh, I got the answer thanks all :)

OpenStudy (experimentx):

what was the answer??

OpenStudy (callisto):

height = 1-2cospi =3 Put y=0 into the equation x=pi/6 or pi = 5pi /6 base = 4pi/6 Area = (4pi/6)(3)(1/2) = pi => that's the answer :)

OpenStudy (experimentx):

looks like my little trick worked.

OpenStudy (kinggeorge):

I derped.

OpenStudy (callisto):

@experimentX Would you mind sharing your little trick with us?

OpenStudy (anonymous):

\(A=\frac{\pi}{6}, B = 3\) so the rectangle has base \(\frac{\pi}{3}\) and height 3 and area \(\pi\)

OpenStudy (anonymous):

1-2cos(2(pi/2)) = 3 0 = 1-2cos(2x) -1/2 = cos(2x) cos(2x) is zero at pi/4 + npi where n is an Integer thus we have roots pi/4 and 5pi/4 so now we just need to subtract them to get the length of the base 5pi/4 - pi/4 = 4pi/4 is the length pi is the length of the base and the height is pi/2 so A = (pi^(2)/2)/2 Area = pi^(2)/4

OpenStudy (experimentx):

same as yours, the max height would be 3(max value), that's obvious now get one point on left by equating to zero, since graph is symmetric, (pi/2 - pi/6)*2 must be base

OpenStudy (anonymous):

lol made a mistake with final answer Area = 3pi/2

OpenStudy (callisto):

@experimentX Then that's not a trick anymore lol

OpenStudy (anonymous):

i will stick with \(\pi\)

OpenStudy (experimentx):

thats how ffm works

OpenStudy (anonymous):

i must have made a mistake in my calculations oh well :L

OpenStudy (anonymous):

could be wrong, but i like \(\pi\)

OpenStudy (callisto):

@satellite73 it's correct :)

OpenStudy (anonymous):

Height = 1-2cos(2(pi/2)) = 3 Base = 5pi/4 - pi/4 = pi A = bh/2 A = 3pi/2

OpenStudy (callisto):

I just found out the answer is right in front of me... the pi pic :P

OpenStudy (anonymous):

calculated the base of the rectangle via \(\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}\) and the height of the rectangle as \(3\) and area as \(3\times \frac{\pi}{3}=\pi\) should have listened to experimentx to begin with

OpenStudy (callisto):

Excuse me.. where is the rectangle?

OpenStudy (anonymous):

how can the base be pi/2 - pi/6

OpenStudy (anonymous):

the rectangle is the two triangles stuck together

OpenStudy (anonymous):

|dw:1334290347591:dw|

OpenStudy (callisto):

Oh...I see... Nice trick @satellite73 very cute solution :)

OpenStudy (anonymous):

ty

OpenStudy (anonymous):

you guys win I'm wrong :)

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