Expert: Paul Klarreich - 12/15/2006

Question
1) Sqroot of x^2 + 4 = 2 Sqroot of 2

2) In a triangle ABC, angle A = 45 degrees and angle B = 60 degrees. The normal from C hits AB  in D, and the normal from B hits AC in E. Furthermore CD = s.
 a) Find the exact expressions for the lengths DB and AD
 b) Find the exact values for the lenghts AC and BC
 c) Find the area of the triangle ABC expressed with s.
 d) Find BE by making two expressions for the area of the triangle ABC equal to eachother.
 e) how big is angle C?
 f) show that sin 75 degrees = Sqroot of 6 + Sqroot of 2 / 4

Hope you can help. If not, please refer me to someone who can. Thanks in advance :-)

Answer
Questioner:  Erik
Category:  Advanced Math
 
Subject:  please help
Question:  

1) Sqroot of x^2 + 4 = 2 Sqrt(2)

2) In a triangle ABC, angle A = 45 degrees and angle B = 60 degrees. The normal from C hits AB  in D, and the normal from B hits AC in E. Furthermore CD = s.
a) Find the exact expressions for the lengths DB and AD
b) Find the exact values for the lenghts AC and BC
c) Find the area of the triangle ABC expressed with s.
d) Find BE by making two expressions for the area of the triangle ABC equal to eachother.
e) how big is angle C?
f) show that sin 75 deg = Sqroot of 6 + Sqroot of 2 / 4

Hope you can help. If not, please refer me to someone who can. Thanks in advance :-)
............................................
Hi, Erik,

1) I'm not sure where you're going with this one, but perhaps this will help:

x^2 = 2 sqrt(2) - 4

Now the right side is negative so you will have imaginary roots:

x = +- i sqrt( 4 - 2 sqrt(2) )

That's actually an answer, but you probably wanted something more elegant, and I'm afraid I can't come up with one.

..............................................
2) In a triangle ABC, angle A = 45 degrees and angle B = 60 degrees. The normal from C hits AB  in D, and the normal from B hits AC in E. Furthermore CD = s.
a) Find the exact expressions for the lengths DB and AD
b) Find the exact values for the lengths AC and BC
c) Find the area of the triangle ABC expressed with s.
d) Find BE by making two expressions for the area of the triangle ABC equal to eachother.
e) how big is angle C?
f) show that sin 75 degrees = Sqroot of 6 + Sqroot of 2 / 4

OK, I think I have your diagram, and I find that (for whatever it's worth):

WARNING: THE FOLLOWING DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

Angle C ( angle ACB) = 75 deg.
Angle BCD = 30 deg
Angle ACD = 45 deg
Angle CBE = 15 deg
Angle ABE = 45 deg.

Also:
In triangle ADC,
CD = s  (given)
AD = s, because ACD is a right isosceles triangle.
AC = s sqrt(2), by the Pythagorean Theorem.

In triangle BCD:

The side opposite 60 = sqrt(3) * the side opposite 30.
But the side opposite 60 is CD = s, so

CD = s = BD sqrt(3)
DB = s/sqrt(3)
CB is the hypotenuse, and is always twice the side opposite 30:

CB = 2 DB = 2s/sqrt(3)

We're getting there.  (said the dentist, as he yanked harder on the tooth.)

I think we have parts (a) and (b) done.

The area of ABC = bh/2

the base is  AD + BD = s + s/sqrt(3) = s(1 + 1/sqrt(3))
      sqrt(3) + 1
AB= s(-------------)
        sqrt(3)

      3 + sqrt(3)
AB = s(-------------), after rationalizing
           3

The height is s.  (That's CD.)

           3 + sqrt(3)
Area = s^2(-------------)
             6
................................

Now how about BE?  You can always take a different base and height.  Base does not mean bottom; any side can be the base, and the height is the 'normal' to that side.  In this case,

Let AC be the base.
and BE be the normal.

AC = s sqrt(2) (we got that earlier)
BE is to be found.

But the area must be the same, so set:

AC * BE / 2 = Area
                      3 + sqrt(3)
s sqrt(2) BE / 2 = s^2(--------------)
                           6

A little simplification and cancelling:
                    3 + sqrt(3)
 sqrt(2) BE     = s(------------)
                         3
      3 + sqrt(3)    
BE = s ------------
      3 sqrt(2)

Rationalize: ( I mean it about that Courier font.)

      3 sqrt(2) + sqrt(6)    
BE = s -------------------
             6
e) Angle C = 75; we answered it above.

f) show that sin 75 degrees = Sqroot of 6 + Sqroot of 2 / 4

Now I have a little problem with this -- I don't know your context.  Ordinarily this is an application of the reduction formula:

sin(A + B) = sin A cos B + cos A sin B,
with A = 45 and B = 30.

But you might not have studied that yet.  So let's try to get it from triangle BEC.

BEC is a right angle, and one of its acute angles is 75.  The leg opposite that is  BE, found in part d.  The hypotenuse is CB.

sin 75 = BE/BC

         3 sqrt(2) + sqrt(6)    
       s -------------------
                6
sin 75 = ------------------------
               2s
             -------
             sqrt(3)

         3 sqrt(2) + sqrt(6)    
         -------------------
                6
sin 75 = ------------------------
               2
             -------
             sqrt(3)


         [3 sqrt(2) + sqrt(6)] sqrt(3)    
sin 75 =  ------------------- -------
                6               2

         3 sqrt(6) + sqrt(3) sqrt(2) sqrt(3)    
sin 75 =  -----------------------------------
                       12

         3 sqrt(6) + 3 sqrt(2)    
sin 75 =  ---------------------
                 12

         sqrt(6) + sqrt(2)    
sin 75 =  ---------------------
                 4

I think that's it. (Sorry it took so long, in more ways than one.)