
Topic review (newest first)
 irspow
 20060123 14:02:54
Take the part you highlighted:
Also, forgive me if I'm wrong but you wrote: A = arcsin[4sin34/5] ≈ 26.574° then The only other possibilities are (180  A) = 153.426
But you didn't highlight:
....but 153.426 + 43 > 180, so this would not be a triangle.
Otherwise, take a look at c) again, as it is one of the data set which describes two possible triangles.
c) A = arcsin[4sin34/2.237] ≈ 89.181 and 90.819
Therefore two triangles can be formed.
Just work it out;
If sinA = asinB/b; sinA = 4sin34/2.237
and A = arcsin(asinB/b), then A = 89.181 is one possiblility
but...180  89.181 = 90.819° is another possibility,
Since 34 + 90.819 < 180, this angle is possible,
Well, 4sin34/2.237 = .99989...
and sin90.819 = .99989...
and sin89.789 = .99989...
This all just means that you can form two different triangles meeting all of the criteria listed in the question.
That seemed long, but I hope it explained it to you. It is just that two different angles can produce the same length side.
 mikau
 20060123 10:07:30
1) Using the law of sines gives;
sinA/a = sinB/b; since nothing varies but b this becomes;
A = arcsin[4sin34/b]
a) A = arcsin[4sin34/5] ≈ 26.574°
The only other possibilities are (180  A) = 153.426
....but 153.426 + 43 > 180, so this would not be a triangle.
b) A = arcsin[4sin34/2] = which does not have a solution
c) A = arcsin[4sin34/2.237] ≈ 89.181 and 90.819
Therefore two triangles can be formed.
d) A = arcsin[4sin34/3] ≈ 48.21 and 131.79
Since 131.79 + 43 < 180 both angles above are possible.
So for this part, data c and d would produce two triangles.
I don't understand this at all, what does it mean by two triangles?
Also, forgive me if I'm wrong but you wrote: A = arcsin[4sin34/5] ≈ 26.574° then The only other possibilities are (180  A) = 153.426
Uh.. thats two angles that sum up to 180. Don't triangles usually have three angles? I'm missing something here. Seems to me, the remaining angle should be 180  A  34.
 irspow
 20060123 08:03:08
1) Using the law of sines gives;
sinA/a = sinB/b; since nothing varies but b this becomes;
A = arcsin[4sin34/b]
a) A = arcsin[4sin34/5] ≈ 26.574°
The only other possibilities are (180  A) = 153.426
....but 153.426 + 43 > 180, so this would not be a triangle.
b) A = arcsin[4sin34/2] = which does not have a solution
c) A = arcsin[4sin34/2.237] ≈ 89.181 and 90.819
Therefore two triangles can be formed.
d) A = arcsin[4sin34/3] ≈ 48.21 and 131.79
Since 131.79 + 43 < 180 both angles above are possible.
So for this part, data c and d would produce two triangles.
2) I assume you know the quadratic equation?
[b ± √(b² 4ac)] / 2a; where a = the constant from the x² term, b = the
constant from the x term, and c = the constant term with no x.
For your equation this becomes;
[7 ± √(49  40)] / 10;
This produces the two roots: 2/5 and 1;
2/5 + 1 = 2/5 + 5/5 = 7/5 or b)
3) The discriminant in the quadratic equation is the part in the radical sign;
±√(b²  4ac)
An irrational number is just a number that cannot be expressed as a exact fraction.
Your choices above would produce;
a) ±√15 = √15 and √15
b) ±√0 = 0 (which is unsigned)
c) ±√16 = 4i and 4i
d) ±√49 = 7 and 7
e) ±√1 = i and i
The only one that represent two irrational numbers is a)
 mikau
 20060123 07:19:26
"For which set of data would 2 triangles be formed?"
The wording of this problem makes no sense at all to me. TWO trianges? What the heck?
The others are easy enough. #2 Just use the quadratic formula to find the roots, then add them and find "the sum of the roots".
3. Well the discriminant is the part of a quadratic formula inside the radical. If the discriminant is negative, the equation would have imaginary roots. So we can ellimante c and e. If it is 0 or 49, the square roots would be integers (7, and 0) and would be rational numbers. The square root of 15 is an irrational number (since it cannot be expressed as a fraction of integers), so a. is the answer.
I need help with these problems:
#1) For which set of data would 2 triangles be formed?(for these questions please show me how do you know the answers please...?) a) <B=34degrees, a=4, b=5 b) <B=34degrees, a=4, b=2 c) <B=34degrees, a=4, b=2.237 d) <B=34degrees, a=4, b=3
#2) The sum of the roots of 5x^2+7x2=0 is a) 7/5 b) 7/5 c) 2/5 d) 2/7
#3) If a quadratic equation has 2 irrational solutions, then its discriminant could equal a) 15 b) 0 c) 16 d) 49 e) 1
