Melody

Here are the steps to the solution:

1)Draw the triangle.

2) Mark in all the angles. You have enough information to find the whole lot just using the angle sum of a triangle.

3) Now just by looking at what kind of triangles you have and you can see the length of AC

4) Use sin rule to find BC    [AC would be ok too but it is easier to get an exact area using BC]

5) Now you have all the angles and 2 sides so use the formula to find the area of ABC.    ( area= 0.5xysinZ,  if it had beeen triangle XYZ)

My answer is different from all the others but I could easily have made a careless error.

If you are really interested in learning then follow my instructions and see where it takes you.  

You can ask questions along the way if you want to.

the question is incomplete.

Ask one question and learn from the answer.

Guest one is right.

We are not here to do your homework for you.

Thanks Justin.

You had basically the right idea too.  We both used complimentary counting.

Here is a very lay definition:

A convergent series is one where the sum of the terms gets closer and closer to a specific constant.  It converges.

A divergent series is one where the sum of the terms diverges off to pos or neg infinity.

If you look at this sight there a a number of examples of divergent and convergent series.

Thanks Justin :)

Are you looking for help Rob

or are you just posting a fun puzzle question?

Either way it is fine but you should make it clear whether or not you are actually looking for help.   

Hi Guest and Justin,

Thanks for answering Justin, just thought I'd give it a go myself.  :)

A debate team consists of 5 freshmen and 4 sophomores.  (Everyone is distinguishable.)  In how many ways can they stand in line, so at least two of the sophomores are standing next to each other, and at least two of the freshmen are at the ends of the line?

The only way you can have  2 freshman at the end and NOT 2 sophomore together is

S F S F S F  S        F F 

So the number of ways this can happen is     5*4  * 3!*4! = 20*6*24 = 2880

Now all the possible combinations wiith  FF at the end but no other restrictions is  5*4 * 7! 

So the number of ways that has 2 freshmen at the end AND at least 2 sofomores next to each other is    

$5*4*7!    \quad  -\quad        5*4  * 3!*4!   \\ =5*4*4!(5*6*7-6)\\ =4*5*4!*6*34\\ =97920$

I suggest you check what we have both done and decide for yourself which answer (if any) is correct.

You ask a question in a new post and if you are lucky, some generous and knowledgable person  answers it.

Then you interact with them or at least say thank you and give them a point.

You are not expected to interact or acknowledge troll answers.   eg "the answer is 5"   Unless you are asking them to expand on their answer..

Thanks for answering Justin.  I just want to have a go at it too  

If all the children are seperated by adults

Place an adult down anywhere there are 2! possibilities for the other 2 adults and 3! for the children

2! * 3! = 2 * 6 = 12 ways to seat them

If 2 of the chilren sit together and 2 adults sit together.

3 ways to chose the two children then 2 ways to seat them. they start the circle so that is 6 ways for them

there are 2 places the 3rd child can go

6*2 so far

then there are 3! =6  ways to seat the adults.

6*2*6 = 72 ways 

Total = 12+72 = 84 ways    [I think]