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I'm preparing for a math competition, and was stumped by this problem. The original problem is shown below, and the correct answer is $120°$. I'm posting this here to ask for explanation on how this answer was reached.

The equation given is $$\tan\theta={{\cos5°\cos20°+\cos35°\cos50°-\sin5°\sin20°-\sin35°\sin50°}\over{\sin5°\cos20°-\sin35°\cos50°+\cos5°\sin20°-\cos35°\sin50°}}$$

I was able to simplify the numerator using $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$, and simplify the denominator using $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$ to get the expression given in the title of this question, but I don't know how to go any further, or if this is even the right first step.

ACB
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Nicholes Cortes
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    Do you know the formulas for $\sin a-\sin b$ and $\cos a+\cos b?$ – Thomas Andrews Mar 21 '23 at 01:40
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    @NicholesCortes Welcome to Math SE. FYI, related to Thomas Andrews' comment above, please see the "Sum-to-product" table in the Product-to-sum and sum-to-product identities section of Wikipedia's "List of trigonometric identities" article. – John Omielan Mar 21 '23 at 01:45
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    $$\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b).$$ $$~\cos(a - b) = \cos(a)\cos(b) +\sin(a)\sin(b).$$ $$\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b).$$ $$\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b).$$ From these identities, you can derive that $$\cos(a) + \cos(b) = 2 ~\cos\left( ~\dfrac{a+b}{2} ~\right) ~\cos\left( ~\dfrac{a-b}{2} ~\right),$$ $$\sin(a) - \sin(b) = 2 ~\sin\left( ~\dfrac{a-b}{2} ~\right) ~\cos\left( ~\dfrac{a+b}{2} ~\right).$$ – user2661923 Mar 21 '23 at 03:33

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We have: $\tan(\theta)=\dfrac{\cos(25^{\circ})+\cos(85^{\circ})}{\sin(25^{\circ}) - \sin(85^{\circ})}=\dfrac{2\cos(55^{\circ})\cos(30^{\circ})}{-2\cos(55^{\circ})\sin(30^{\circ})}=-\cot(30^{\circ})=-\sqrt{3} \implies \theta = 120^{\circ}$ would be the smallest positive angles satisfy the equation.

Wang YeFei
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