# Thread: Somewhat basic Cosine rule problem. Help appreciated

1. ## 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)

Click the image to make it larger.

2. ## 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.

3. ## Re: Somewhat basic Cosine rule problem. Help appreciated

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?

4. ## 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.

5. ## Re: Somewhat basic Cosine rule problem. Help appreciated

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.

6. ## Re: Somewhat basic Cosine rule problem. Help appreciated

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,

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

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