Imageshack - 87978225.jpg thats the problem please help i have a test tomorrow and if i fail i will loose one year please help

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- Aug 18th 2010, 09:14 AMTheGameTrigonometry problem , easy for someone who knows math :)
Imageshack - 87978225.jpg thats the problem please help i have a test tomorrow and if i fail i will loose one year please help

- Aug 18th 2010, 05:34 PMAckbeet
Is this problem for studying for the test? Or is it a take-home test?

- Aug 20th 2010, 04:15 AMumangaroraAnswer
Alpha=x.. k?

cos x=5/13

x is between 270,360 so x is negative of the form (360-x).

cos x= cos -x

therefore, we can take cos x as acute.

but sin x is negative

sin360-x = - sin x

sin(60-x)=sin60*cosX -sin(-X)cos60

=root3 *5/13 + 12/13*1/2

=5root3 + 12

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26

cos(45+x)=cos45cosX-sin45sin(-X)

=1/root2*5/13 + 1/root2*12/13

= 17

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13root2 - Aug 20th 2010, 04:43 AMAckbeet
Your addition of angles formula is the right way to go about this. The restriction on x just tells you the sign of sin(x), where you use Pythagorean theorem to find sin(x) without actually needing to solve for x. I would double-check your trig there, especially the sin(60).

- Aug 20th 2010, 06:54 AMSoroban
Hello, TheGame!

Quote:

Find: .

. . . . .

is in a right triangle with:

Pythagorus tells us that:

Since is in

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . .

- Aug 20th 2010, 11:25 AMumangarora
Also,i use sin^2 x + cos^2 x=1.. another form of pythagoras.. but straight to the point..