Alan

Are you looking to turn rectangular coordinates into polar coordinates?

If so, pol(x,y) becomes (r, θ) where θ = tan^-1(y/x) and r = √(x² + y²)

However, I don't know if the values you've supplied are meant to indicate the end values of a range, or if the numbers are to be subtracted or what!  

You are right heureka;  sin is also positive in the second quadrant, where cos is negative.

Do you mean find p?  If so then p+10 = 90 - 49, so p = 31°.  Another possibility is in the second quadrant where p + 10 = 180 - 41 so p = 129°  (The 41 comes from the first p+10, i.e. 31+10 = 41), 

Check:

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{49}}^\circ\right)} = {\mathtt{0.656\: \!059\: \!028\: \!991}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{31}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}\right)} = {\mathtt{0.656\: \!059\: \!028\: \!991}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{129}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}}\right)} = {\mathtt{0.656\: \!059\: \!028\: \!991}}$$

Could also note that  cos(2θ) = 2cos²(θ)-1 so 1-2cos²(60) = -cos(120)

$${\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{120}}^\circ\right)} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0.5}}$$

No need to think of removing Chris's points!  There are 10 possible choices for the first wheel.  For each of these 10 there are 9 possible choices for the next wheel.  For each of these 10*9 there are 8 possible choices for the next wheel.  For each of these 10*9*8 there are 7 possible choices for the last wheel.  

$${\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{7}} = {\mathtt{5\,040}}$$

What, even with those shades!!!

For the general quadratic equation a*x² + b*x +c = 0 the solutions are given by:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$  

By rearranging 7x² = 2x - 5 to 7x² - 2x + 5 = 0 we have here that a=7, b=-2, c=5 so

$$x=\frac{2\pm\sqrt{2^2-4*7*5}}{2*7}$$  which results in:

$${\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{34}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{7}}}}\\ {\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{34}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{7}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{7}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.832\: \!993\: \!127\: \!834\: \!905\: \!8}}{i}\right)\\ {\mathtt{x}} = {\frac{{\mathtt{1}}}{{\mathtt{7}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.832\: \!993\: \!127\: \!834\: \!905\: \!8}}{i}\\ \end{array} \right\}$$

If sin=4/5 then we can picture the well known 3, 4, 5 right-angled triangle; so cos = 3/5.

If his anger doubles every 20 minutes, then, because there are 9 lots of 20 minutes in 3 hours his anger will have increased by 2⁹ or 512 times!

 I'll expand on the above a little:

After 20 mins (1*20mins) his anger doubles, i.e. increases by 2¹

After 40 mins (2*20mins) his anger has doubled again, total increase = 2¹*2¹ = 2² 

After 60 mins (3*20mins) his anger has doubled again, total increase = 2²*2¹ = 2³ 

After 1 hour 20 mins (4*20mins) ... , total increase = 2³*2¹ = 2⁴ 

... etc.

After 3 hours (9*20mins) ..., total increase = 2⁸*2¹ = 2⁹. 

Can't trust Google translate, so: