Tuesday, June 10, 2008

Special Angles in Trigonometry - Part 2 (Similar Triangles)

This post will hopefully clarify how to work with special triangles when the sides are not the standard lengths (as I described in the original post for Special Angles).

Maybe this would be a good time to describe SIMILAR triangles. Similar triangles are triangles that have different side lengths, but have the same angles. Don't let the fancy name fool you... just think of them as smaller or larger versions of the same triangle. Examine this picture:

If you were to cut out all the triangles, and shrink or enlarge them, you would see that they all would fit on top of each other. This is possible because each triangle has the same angles, despite having different side lengths. When triangles are like this, they are said to be SIMILAR.

Now, to apply this to my previous post explaining the special angles in trigonometry. I explained how to derive the trig functions using the simplest triangles. However, in all likelihood, you will find triangles of different sizes, rather than these same simple triangles. You need only remember the rules of SOHCAHTOA to be able to evaulate the trig functions.

For example:

Take a right angle triangle with an angle of 30 degrees, and you know that the short side is a length of, say, 5. (Try to sketch this out... being able to draw a triangle from a description is important to learn as well!) There are a few strategies you could use here to solve the other 2 sides and last angle.

1) You know that all the angles in a triangle sum up to 180 degrees. So, 180-90-30 = 60!

2) You can apply the rules of SOHCAHTOA to determine either one of both of the unknown sides. In this case, you have the 30 degree angle known, and the short side is opposite this angle. So, you can use the Sine function to determine the length of the hypotenuse (try it! how do you solve Sin(30). Once you have the hypotenuse figured out you can then turn around and use it to to solve the Cosine function... OR you could just use the Theorem of Pythagoras, since it is a right-angle triangle.

3) Remember that you could also use either the Sine Law or Cosine Law in there as well (especially on triangles that are NOT right-angle triangles)... Sine Law works whenever you know an angle and it's opposite side, and then either of an angle or a side to complete the identity. Or the Cosine Law will work when you know 2 sides and the angle between those 2 sides.

Hopefully this quick post addresses some concerns that some of you may have had. Sorry about any confusion or lack of clarity in the first post. As always, please continue to drop me a line if you have any questions, concerns, or topics you would like me to cover!

**EDIT** - My apologies... as pointed out in the comment section, triangles as I have discussed in this post are in fact called SIMILAR, and not CONGRUENT (I have edited the post to reflect this). Similar triangles are triangles with the same angles but can have differing side lengths. Congruent triangles, on the other hand, have the same angles AND sides. This means they look either the exact same, or are a mirror image of the original. Sorry for the confusion, and thanks to the astute reader who picked out my error. :)

Tuesday, March 11, 2008

Word Problems - Hints and Tips to Solving Them

Word problems are one kind of math question that ALWAYS gives students troubles, probably because making a connection between the math and the 'real world' situation isn't necessarily the easiest thing to do. Despite that, there are a few pointers that you can try to which will hopefully lead you to the correct solution.



Joe wants to save money to buy a new Playstation 3. It costs $400 to buy from the store. He has a job which pays him $15 per hour, and he works 6 hours per day. However, to get to and from his job, he has to take the bus, which costs him $2 each way, and he buys lunch each day he works for $6. He also owes his dad $500 from when he borrowed it to buy a new stereo. Taking all of these costs into account, how many days will it take Joe to save up enough money to buy his new Playstation? How much more money does Joe need to earn if he wants to buy it in 5 days?

PS3 cost = $400
Wages = $15 / hr
Work day = 6 hr / day
Bus fare = 2 x $2 / trip
Lunch = $6 / day
Owes Dad = $500

Next, it is important to IDENTIFY WHAT YOU WANT TO FIND OUT. In this case, we want to find out how many days it will take Joe to save enough to buy his Playstation. Specifically, we want to know how long it will take him to save $400.

The next pointer is to try to CLASSIFY WHAT YOU ARE GIVEN INTO GROUPS. In this question, you can classify things into "Things Related to Making Money" and "Expenses."

Wages = $15 / hr
Work day = 6 hr / day
Bus fare = 2 x $2 / trip
Lunch = $6 / day
Owes Dad = $500
From here, it might start becoming more apparent what need to be done. In this case, you would find out how much money Joe makes in a day, and also how much he spends in a day.
Earnings = $15 / hr x 6 hr / day = $90 / day
Expenses = 2 x ($2 / trip) + $6 / day = $10 / day
Overall = 90 - 10 = $80
So, Joe saves $80 per day.
Now, he owes his Dad $500, and wants to buy his Playstation for $400... totalling $900. At $80/day, Joes can pay off his Dad and then be able to afford his Playstation in:
$900 / ($80/day) = 11.25 days.... = 12 days (since he can't stop working at 11.25 days). :)
Of course, now that you have an answer, make sure that you answer the question completely.... there's still another part!
If Joe wants to make his purchase, after paying back his dad, in 5 days.....
total cost = $900
time = 5 days
savings per day = ?
When you separate what you know from what you don't know, it is very helpful in seeing how to get to the answer. In this case, Joe needs $900 / 5 days = $180. Compared to what he makes right now, he needs to save another $100 per day!
So to summarize, they keys to word problems are to:
IDENTIFY WHAT YOU KNOW (classifying into groups helps here),
Simple rules, but important nonetheless. :)

Wednesday, February 27, 2008

Word Problems and Unit Conversions

For some people, working with variables that have different units is very easy. They can conceptualize the relationship between the different variables in a way that seems natural to them, yet remains very difficult to explain to others. There are even more people who cannot do this type of math so automatically, and often struggle to solve problems... myself included. However, there is a trick to solving these questions, and the solution lies, quite simply, in keeping track of the units. I will try to demonstrate.

I will start simply, with multiplication and division. You know that if one number is on the top and on the bottom of a division line, you can cancel them out to reduce things:

You also know that you can do this when you are working with variables, like x and y:

So then, it's only natural that you can do this with units as well!

Now, here is the trick. If you are struggling with the concept of a word or other problem that deals with units, change the way that you are looking at it, to focus on the units involved. They will guide you towards the final solution.

Let's try a basic physics problem: How far will an object travel in 10 seconds if it is moving at a constant speed of 25 meters/second? This one might not be too difficult to conceptualize, but it demonstrates my point. Watch what I do with the units at the start, if you are not familiar with this part.

As you can see, the units cancel out to leave just the 'meters.'

Let's try one that is a bit more complicated: If a force of 10 N pushes a 10 kg block along a flat, frictionless surface for 5 seconds, what is the block's average velocity? This one is undoubtedly more difficult to do in your head, but if you keep track of the units (what you have, and what you need), then you should be able to do it. For those who aren't in physics, it is easier to know that the unit Newton, N, is equal to (kg) x (m) / (s)^2... blogger is still terrible with superscripts and equations. :(

As you can see, we wanted velocity (m/s) in the end, and so I just worked with the units until I got what I wanted. From a physics point of view, I used Force = Mass x Acceleration, and Velocity = Acceleration x Time.
Currency conversions are a common everyday example of where you can use this trick. If I say that the exchange rate is 0.985 $US / $CAN, and I want to know how much $US I can exchange for $232 CAN:

Naturally, it's best to try to understand the concepts of the questions, and the more complex questions won't be quite so easy to muddle through just by tracking the units. However, hopefully you can see the benefit of looking at these types of problems from this angle, and it will help you to solve them. I learned this technique in high school, and I still benefit from it. As always, feel free to drop me a line if you would like me to clear anything up. :)

Tuesday, February 12, 2008


I have a request to go over polynomials, and to specifically explain what FOIL means.

"Polynomials" refer to mathematical expressions that contain multiple terms (poly = multiple, nomials = terms/numbers (roughly)). "Monomial" refers to a polynomial that has only a single (mono) term. Furthermore, the product of multiplying monomials together also results in a monomial. Basically, think of it along these lines:

- 1 term (bunch of things multiplied/divided/raised to a power) = monomial

- more than 1 term (monomials) that are added or substracted = polynomial

Some examples hopefully clear this up:

1) Monomials:

2) Polynomials:
(5xyz^3)/4 + 7x

Doing calculations with polynomials is fairly straight-forward, but you do have to keep a few things in mind.

For addition or subtraction of polynomials, you must remember that you can only combine 'like terms.' 'Like terms' contain the sample variables and differ by the numeric coefficient in the front. 3x and 5x are like terms, and can be added to get 8x. However, 3x and x^2 are not like terms and CANNOT be added like in the first example.

x + 2x = 3x
4x + 4x = 8x
15x + 100x = 115x
50x - 10x = 40x

You cannot add terms in this way that have different variables:
x^2 + 5x
3x^4 - 2x^3
x + 7

**Technically, some expressions such as these can be SIMPLIFIED by pulling out a common factor, but they still cannot be added up, as with like terms:
eg. x^2 + 5x = x(x + 5)

Multiplying polynomials is a little more complicated, but still straight-forward, provided that you keep track of what you are doing. When multiplying two polynomials together, all that you need to do is add up the products of multiplying each term in the first polynomial by each term in the second polynomial. You have probably heard the expression "FOIL" when talking about multiplying polynomials. FOIL is short for "First, Outside, Inside, Last" and refers to which terms you multiply together and add up when multiplying two polynomials, each composed of two monomials (specifically, this is the product of two BINOMIALS).

Here is an example of this:

(2x + 3)(x + 5) = ?

First: 2x * x
Outside: 2x * 5
Inside: 3 * x
Last: 3 * 5

Adding these up gives:
2x^2 + 10x + 3x + 15

Furthermore, as I showed earlier, about adding polynomials with like terms, you can simplify this expression:
2x^2 + 13x + 15

And that's the final, reduced answer. You can apply the FOIL principle to any two binomials to arrive at their product. When you have more complicated polynomials, such as those composed of 3, 4, 5, or more monomials, you do the same type of thing... what I find easiest is to take the first term of the first polynomial, and multiply it with every term of the second polynomial. Then do the same for the second term in the first one, multiplying with every term in the second one, and so on.

Like this:
(x^2 + x + 5)(x^3 + x^2 + 1)

First group (multiply x^2 with all in the second polynomial)
x^2 * x^3
x^2 * x^2
x^2 * 1

Second group (x)
x * x^3
x * x^2
x * 1

Third group (5)
5 * x^3
5 * x^2
5 * 1

Now, you just add up all these terms, and simplify where you can:
(x^2 * x^3) + (x^2 * x^2) + (x^2 * 1) + (x * x^3) + (x * x^2) + (x * 1) + (5 * x^3) + (5 * x^2) + (5 * 1)

(x^5) + (x^4) + (x^2) + (x^4) + (x^3) + (x) + (5x^3) + (5x^2) + (5)

(x^5) + 2(x^4) + 6(x^3) + 6(x^2) + x + 5

And that's it. A little more complicated, but as long as you keep track of what you're doing and work your way through it, you will arrive at the answer!

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