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Category: 12th Grade

Solve the Triangle. Round Angle Measures to the Nearest Minute & Side Measures to the Nearest Tenth. a=11.4, b=13.7, c=12.2

Monday, September 6th, 2010


These are the formulas used to solve triangles:

1. The sum of the internal angles equals 180o …A + B + C = 180o

2. The ‘sine rule’ …

3. The ‘cosine rule’ …

a² = b² + c² – 2bc cosA


b² = a² + c² – 2ac cosB


c² = b² + a² – 2ba cosC

In the problem that you gave us, we know the length of the three sides of the triangle:

a = 11.4

b = 13.7

c = 12.2

When no angles are known, the cosine rule is the only option, so 11.42 =13.72 +12.22 – 2*13.7*12.2*cosA….Therefore cosA = 0.617955.

You can now use a calculator or a table to find the value of A. You should get A = 0.9047 radians = 51.833 degrees = 51 degrees and 50 minutes.

Use the sine rule to find one of the remaining angles.


So sin(B) =  = 0.944835

You can now use a calculator or a table to find the value of B. You should get B = 1.237 radians = 70.8801 degrees = 70 degrees and 53 minutes.

Finding the third angle is easy, since we know that A + B +C =180 degrees. C = 180 – 51.833 – 70.880 = 57.287 degrees = 57 degrees and 17 minutes.

Visual representation:

What is the Largest Amount of Postage That Cannot be Made Using These Stamp Denominations: 17 Cents, 9 Cents & 5 Cents?

Monday, September 6th, 2010


We’re not sure we understand the question, because if you have enough stamps (of any denomination) you can cross any amount. However, you said that the total value (let’s call is “q”) has to be less than a million dollars (e.g. $999,999.99), so let’s work backwards.  $999,999.99 can be obtained as follows:

  1. 5,882,350 stamps of 17 cent denomination
  2. 1 stamp of 9 cent denomination
  3. 8 stamps of 5 cent denomination

So we have shown the largest amount of postage that can be paid (below the million dollar limit) is $999,999.99.

Now let’s assume that the amount paid has to be exact, i.e. we cannot overpay. This is a lot more complicated, as to do this we have to consider all values of q, where q = 17x + 9y + 5z, and where x, y and z are positive integers.  We could write a computer program to check the value of q for all values of x, y and z, and then see which values are missing, and finally pick the highest number among the missing values. We might want to work backwards from $999,999.99 (since we know we can reach $999,999.99) and check to see if the a small amount is possible. For example, we know that we can remove one of the 8 stamps of 5 cent denomination, so that gets us to $999,999.94. We could remove another 5 cent stamp to get to $999,999.89, and then add a 9 cent stamp to get to $999,999.98 (x= 5,882,352; y=2; z=6). We can continue this exercise until we hit a number that we cannot create with any combination of values for x, z and z.

The easiest way to solve the problem is to recognize to work backwards to prove that we can calculate most values of q below $1,000,000:

  1. We have defined q=17x+9y+5z
  2. We know that q-1=17(x)+9(y+1)+5(z-2)
  3. So let’s start with (x=0; y=0, z=20,000,000), whereby q=$1,000,000
  4. Now increase y by 1 and decrease z by 2: q=9,999,999.99
  5. Again increase y by 1 and decrease z by 2: q=9,999,999.98
  6. Continue this until z reaches zero (x=0; y=10,000,000; z=0), where  q=$900,000
  7. Start again with (x=0;y=0;z=18,000,000), where q=$900,000
  8. Repeat the process of decreasing 1 y by one and decreasing z by 2 (you can do this in an Excel spreadsheet)
  9. Continue until z reaches zero, and then reset y to zero and z to whatever is required to reach the last value of q (or a number slightly higher)
  10. Repeat the exercise until you reach a value of q = 32 cents.
  11. Now you need the 17 cent stamp to get to q=31 (x=1,y=1,z=1).
  12. You will also need the 17 cent stamp for q=26 and q=22 (see blow)
  13. You can find solutions for q down to 22, but you will get stuck on the number 21…so that must be the answer!

Cos 2x=0.32

Monday, September 6th, 2010


First let’s apply the reverse cosine function to both sides of the equation:

cos-1(cos(2x)) = cos-1 (0.32) = cos-1 ()

Now if you take the inverse cosine the  cosine of 2x, you get 2x…so:

2x = cos-1 ()

2x 71.34°

x 35.67°  (which is approximately the same as 0.622 radians)

This looks like it might be the final answer, but actually it’s only one of the many correct answers. One way to see this is by graphing y=cos(2x) and seeing where it intersects the line y=0.32. If you do this, you will see that the intersections occur at multiple points:

As you can see, the intersections occur at x 0.622 radians, 2.519 radians, 3.764 radians, 5.660 radians, etc. The general form of the solution set is as follows:

Find the critical numbers of f (x) = x^2-6x. Find also the open intervals on which the function is increasing or decreasing and locate all relative extrema.

Saturday, July 24th, 2010


A number a in the domain of a given function f is called a critical number  ‘(a) = 0 or f ‘ is undefined at x = a.

Relative extrema are the minimums or maximum points on a part of a curve, while absolute extrema are the minimums and maximum points along the entire curve.

Now we are looking at the following curve:

f (x) = 

To find the extrema, we take the first derivative and figure out at what value of x, the first derivate of the function equals to zero:

f’(x) = 2x-6

f’(3) = 0

As you can see the function is decreasing from negative infinity to 3, and increasing from 3 to infinity. The critical number is 3. There is only one relative (or local) minimum at (3, -9).

Use a half angle identity to find the exact value- sin(75 degrees)

Friday, July 23rd, 2010


The line of a half angle identity is as follows:

If we plug in 75 degrees we get:

A particle is trapped in an infinite one-dimensional well of width 0.132nm. The electron is in the n=10 state. A) What is energy of the electron? B) What is the uncertainty in its momentum?

Friday, July 23rd, 2010