I need to get back in the groove

[sh86]

Hi everyone. Let me start by saying that my math is quite rusty to say the least. I got through high school math through memorizing formulas and as a result I have large, gaping holes in my understanding of mathematics and I'm trying desperately to make up for it. This is especially true for my knowledge of geometry and parts of trigonometry.

Take a look at this math problem page I found on Google: http://www.coastal.edu/math/mathcontest/Level1_04.pdf . See number 3? How do you solve this problem? I have no clue where to begin. I don't remember the strategies. I have the same issue with number 7. Number 10 is a mystery as well (notice these all have to do with angles and triangles...).

Exactly what kind of math is used to solve those problems? Is it straight geometry? Trigonometry? General precalculus? And do you know of any good books I can use to teach myself these things? I am looking for a regular math book, not a scholastic textbook used in a classroom. I have also seen some SAT test questions involving angles like this and I am stumped on them too. I do not need help remembering what a logarithm is or how to factor quadratic equations--I remember all that and find it extremely easy. I am just stumped on the types of problems mentioned above. If you know of some good books or a strategy I should follow then I would greatly appreciate your help. The fact that I cannot solve these problems but at one time successfully completed a calculus course (with an A) is very depressing. It literally keeps me up at night.

Thankyou in advance to anyone that replies.

 

[HallsofIvy]

In number 3, you have a right triangle which is half of an equilateral triangle (since the angles in any triangle must add to 180, if they are all equal they must be 60 degrees which is the angle given). That means that if we take the hypotenuse of that right triangle to be "x", the leg adjacent to the 60 degree angle is x/2. Calling the third side "y", by the Pythagorean we have (x/2)²+ y²= x² so y²= x²- x²/4= (3/4)x² and y= (√(3)/2)x. The ratio of those two sides, RO to OQ is (x/2)/y= 1/√(3)= √(3)/3, choice E. Of course, MO and OQ are the same- they are both radii of the same circle.
If you know trigonometry you could do that a little faster by using the fact that sin(60)= √(3)/2.

 

[sh86]

Thankyou HallsofIvy. The solution to that problem was much shorter than I thought it would be. I suppose now I'll go try harder on problem 7. I've gone a few steps into it so far but I think there's some triangle law I need to use to continue that I'm not aware of...

But I'm curious, do you (or anyone else) know of a good book I can read to relearn this stuff? I really want to find a book that can teach me to solve problems such as number 7 in my first post. If anyone could point me in the direction of such a book then that would be really great.

 

[Ouabache]

On #3, I used the later approach (trig) that HallsofIvy mentioned. I noticed that the side opposite to 60deg angle is QO and the side adjacent to same angle is RO. From trigonometry, the tangent (angle) = length of opposite side / length of adjacent side.
tan (60deg)= QO/RO.
I also noticed QO = MO,
therefore ---> tan (60deg) = MO/RO

The question asks, what is the length of RO/MO?
Looking at the above, we see, it is 1/[tan(60deg)]
If you recalled the trig values of common angles along the unit circle, you would have the cos(60deg) = 1/2 and sin(60deg) = √(3)/2.
property of tangent: tan = sin/cos,
so for this angle --> tan(60deg)= [√(3)/2]/[1/2]
we want the inverse 1/[tan(60deg)] = [1/2]/[√(3)/2] = 1/√(3) = √(3)/3 (Ans = E)
If you recalled trig function, cotangent you could have arrived at sol'n a little faster.

 

[Ouabache]

For help on #7, you probably want to do a review of geometry.
If you are just rusty on these concepts, this website can get you going ---> http://www.mathleague.com/help/geometry/geometry.htm