Math Is Fun Forum

katy

Member

Registered: 2005-12-28

Posts: 14

help me please.!

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+7x-2=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

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: help me please.!

"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.

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A logarithm is just a misspelled algorithm.

irspow

Member

Registered: 2005-11-24

Posts: 1,055

Re: help me please.!

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)

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I am at an age where I have forgotten more than I remember, but I still pretend to know it all.

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: help me please.!

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.

Last edited by mikau (2006-01-22 11:09:29)

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A logarithm is just a misspelled algorithm.

irspow

Member

Registered: 2005-11-24

Posts: 1,055

Re: help me please.!

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.

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I am at an age where I have forgotten more than I remember, but I still pretend to know it all.