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Somewhat basic Cosine rule problem. Help appreciated

First part of the question was easy,

The second part is where i'm stuck.

What I have tried :

-Make d the subject in both cases and have one equal to each other (All that happens is that everything cancels each other out and I get 0=0)

-Make d^2 the subject (I don't know why I tried this, same result as before >.>)

Completely lost as to how to solve this problem.

Any help would be appreciated!

(Sorry for poor diagram etc. in advance)Attachment 23458

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Re: Somewhat basic Cosine rule problem. Help appreciated

Because alpha +beta=180 then cos(alpha)=-cos(beta), so you can make an equation using results from the first part.

Re: Somewhat basic Cosine rule problem. Help appreciated

Quote:

Originally Posted by

**biffboy** **Because alpha +beta=180 then cos(alpha)=-cos(beta**), so you can make an equation using results from the first part.

Could you please explain this part to me?

Thanks for the reply ^_^

Re: Somewhat basic Cosine rule problem. Help appreciated

When 2 angles add up to 180 they have the same numerical value but one is the negative of the other eg cos120 =-cos60, cos110=-cos70 Check some values on your calculator.

Re: Somewhat basic Cosine rule problem. Help appreciated

Quote:

Originally Posted by

**biffboy** When 2 angles add up to 180 they have the same numerical value but one is the negative of the other eg cos120 =-cos60, cos110=-cos70 Check some values on your calculator.

Thanks, I understand completely now.

Re: Somewhat basic Cosine rule problem. Help appreciated

Quote:

Originally Posted by

**biffboy** When 2 angles add up to 180 they have the same numerical value but one is the negative of the other eg cos120 =-cos60, cos110=-cos70 Check some values on your calculator.

That's only true for the Cosine and Tangent functions and their reciprocals.

$\displaystyle \cos \(180-\theta)=-\cos \theta$

$\displaystyle \sec\(180-\theta)=-\sec\theta$

$\displaystyle \tan\(180-\theta)=-\tan\theta$

$\displaystyle \cot\(180-\theta)=-\cot\theta$

**However,**

$\displaystyle \sin(180-\theta)=+\sin \theta$

$\displaystyle \csc(180-\theta)=+\csc\theta$