SOLUTION: \overline{AB} and \overline{AC} are legs of isosceles \triangle ABC. Point $D$ lies on \overline{AB} such that AD = 3, CD = 4, and CB = 5. What is \angle A in degrees?

Algebra ->  Geometry-proofs -> SOLUTION: \overline{AB} and \overline{AC} are legs of isosceles \triangle ABC. Point $D$ lies on \overline{AB} such that AD = 3, CD = 4, and CB = 5. What is \angle A in degrees?      Log On


   



Question 1210686: \overline{AB} and \overline{AC} are legs of isosceles \triangle ABC. Point $D$ lies on \overline{AB} such that AD = 3, CD = 4, and CB = 5. What is \angle A in degrees?
Found 2 solutions by ikleyn, n2:
Answer by ikleyn(54005) About Me  (Show Source):
You can put this solution on YOUR website!
.
AB and AC are legs of isosceles triangle ABC. Point D lies on AB such that AD = 3, CD = 4, and CB = 5.
What is angle A in degrees?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        Such a triangle, as described in the post,  DOES  NOT  exist,  at all.
        And can not exist.
        The proof is below.


Let x be the side leg of the isosceles triangle ABC:   x = AB = AC.


Write the cosine law for triangle ABC

    x^2 + x^2 - 2*x*x*cos(A) = 5^2,

or

    2x^2 - 2x^2*cos(A) = 25,

    2x^2*(1-cos(A)) = 25.      (1)


Next, consider triangle ADC and write the cosine law for it

    3^2 + x^2 - 2*3*x*cos(A) = 4^2,

or

    x^2 - 6x*cos(A) = 7.       (2)


From (1), express cos(A) via x

    1 - cos(A) = 25%2F%282x%5E2%29  --->  cos(A) = 1 - 25%2F%282x%5E2%29 = %282x%5E2-25%29%2F%282x%5E2%29.   (3)


From (2), express cos(A) via x

    cos(A) = %28x%5E2-7%29%2F6x.         (4)


From (3) and (4)

    %282x%5E2-25%29%2F%282x%5E2%29 = %28x%5E2-7%29%2F%286x%29.     (5)


Since x can not be zero, we can remove the common factor 2x from the denominators in (5).


Then, simplifying, we get

    3*(2x^2-25) = x*(x^2-7),

    x^3 - 6x^2 - 7x + 75 = 0.


This cubic equation has only one real root, and this root is a negative number x = -3.2507

    ( Wolfram Alpha online calculator 
         https://www.wolframalpha.com/input?i2d=true&i=Power%5Bx%2C3+%5D-+6Power%5Bx%2C2%5D-7x+%2B+75+%3D+0
      )


Since there are no positive real roots, it means that a triangle, as described in the post, does not exist.

At this point, my solution is complete.



Answer by n2(95) About Me  (Show Source):
You can put this solution on YOUR website!
.

Today’s posts demonstrate the utter sloppiness and absolute mathematical illiteracy
of the person who devised or created the problems submitted today.

Such people must be ruthlessly expelled from the educational process
and kept at a distance of at least a cannon shot.

Dear visitor, you have already posted enough gibberish on this forum today to get yourself blacklisted.

One more day of such posts, and I will write an email to the managers of this project
asking them to remove you from your duties on the project, since you are not able
to provide correct/proper input.