Lesson A nice Geometry problem solved using Algebra

A nice Geometry problem solved using Algebra

Problem 1

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


Write the cosine law for triangle ABC

    AB^2 + AC^2 - 2*AB*AC*cos(A) = CD^2,

    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

    AD^2 + AC^2 - 2*AD*AC*cos(A) = CD^2,

    3^2 + x^2 - 2*3*x*cos(A) = 5^2,

or

    x^2 - 6x*cos(A) = 16.       (2)


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

    1 - cos(A) =   --->  cos(A) = 1 -  = .   (3)


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

    cos(A) = .         (4)


From (3) and (4)

     = .     (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^2-16),

    x^3 - 6x^2 - 16x + 75 = 0.


This cubic equation has three real roots: they are

    x = 3,  x =  -  = -3.72015,  x =  +  = 6.72015.

    ( see Wolfram Alpha online calculator 
          https://www.wolframalpha.com/input?i2d=true&i=Power%5Bx%2C3%5D+-+6Power%5Bx%2C2%5D+-+16x+%2B+75+%3D+0
      )

The negative root does not fit, so we discard it.

The root x = 3 also does not fit, so we discard it, too.


The only root which works is  x =  +  = 6.72015, approximately.


With this x, let's calculate cos(A) using formula (4). It gives cos(A) = 0.72321, approximately.


So, A = arccos(0.72321) = 0.76236 radiance, or about 43.68 degrees.


ANSWER.  Angle A is about 43.68°.