Find angle of a right triangle. Ok, so this question is from a practice exam. Looks very simple and basic, but I'm not very good at math, so I'm having trouble setting up the problem. 
One acute angle of a right triangle is not less than 30° more than twice the other acute angle. What are the possible measures of the larger angle? 
I know that the sum of all angles equal to 180° and that one side equals 90° so there is a 90° left for the other two remaining angles but I don't know how to divide it. To be honest I don't  I understand the problem where it says "less than 30° more than twice" if someone can put it in different words or explain. I need a hint. Thanks
 A: You're off to a good start! Some people learn better geometrically, however I'll do this algebraically. Like you said, we have two angles we're trying to find the "size" of. Let's call them $\angle a$ and $\angle b$. You've already pointed out that
$$\angle a+\angle b=90^\circ$$
Now, one of these angles must be smaller than the other (or the same size if we have an isosceles triangle), so let's assume that $\angle a$ is the smaller of the two. But notice then that $\angle a\le45^\circ$, as if $\angle a>45^\circ$ then $\angle a$ is bigger than $\angle b$ (can you see why?)
Now, we decipher the actual question. It says that one of the acute angles is not less than $30^\circ$ plus twice the other angle. To say that one angle is not less than another is actually the same as saying it is greater or equal to that other angle. But this means that the bigger angle must be greater than or equal to $30^\circ$ and twice the other, smaller angle (ask yourself, is it clear why it is the bigger angle that is greater than the other expression?). In our notation, this means that
$$\angle b\ge2\angle a+30^\circ$$
Remember! $\angle a+\angle b=90^\circ$, so $\angle a=90^\circ-\angle b$. Substituting this in gives
$$\angle b\ge2(90^\circ-\angle b)+30^\circ$=180^\circ-2\angle b+30^\circ$$
Moving the angle over to the other side of the equality gives:
$$3\angle b\ge180^\circ+30^\circ=210^\circ$$
so
$$3\angle b\ge 210^\circ$$
I'll leave the division to you. The only thing left to consider is just  how big can the larger of the two angles get? (Hint: could it be larger than $90^\circ$? Could it even be $90^\circ$?)
A: $$A+B+C=180°$$
$$B=90°$$
$$A>=2C+30$$
Hence,
$$A+C=90°$$
$$A>=2C+30°$$
Let us set $A=2C+30°$, getting $2C+30°+C=90°$, so $C=20°$ and $A=70°$.
If $C$ is decreased then $A$, will still meet the equals below "Hence," so $A>=70°$.
