Trisecting a 90 Degree Angle?

[geekily]

"Create an angle of 30º using only the figure below."

Figure is a triangle with angles of 40, 90, and 50 degrees. I bisected the 90 to get 45... not sure why, I thought that'd get me somewhere. I filled in the bottom angles that resulted as 95 and 85 degrees. I know there must be some way to trisect the 90, but I can't figure it out. It may have something to do with the fact that it's 2:30 in the morning. (No, it's not that I procrastinated - I've worked on the same assignment for about 10 hours over the last 2 days.) Anyway... if you could help, I'd really appreciate it. Thanks!

 

[Denis]

You know how to construct a 30-60-90 triangle, right?
So construct one and "lay it properly" inside your 40-50-90 triangle: get it?

 

[TchrWill]

"Create an angle of 30º using only the figure below."

Figure is a triangle with angles of 40, 90, and 50 degrees.
.A
.*
.*..*
.*.....*
.*........*....*F
.*...........*
.*..............*
.*........*........*.......*
B.........D........C.......E

Angle ABC = 90º
Bisect BC at D
With radius CD, swing an upper arc from D to E on BC extended
Draw line from B to point F, tangentcy point on 180º arc DFE.
Angle FBC = 30º as sin FBC = CF/BC = 1/2.
Now bisect angle ABF and you have the trisected 90º angle.

 

[TchrWill]

Alternatively:

"Create an angle of 30º using only the figure below."

Figure is a triangle with angles of 40, 90, and 50 degrees.
.A
.*
.*..*
.*.....*
.*........*....*F
.*...........*
.*..............*
.*........*........*.......*
B.........D........C.......E

Angle ABC = 90º
Bisect BC at D
With radius CD, swing an upper arc from D to E on BC extended, 180º
Using BC as radius, swing two arcs from B and C intersecting at G above BC.
Triangle BCG is equilateral making angle GCB = 60º
Draw line CG intersecting 180º arc at F.
Draw line from B to point F
Angle GCB = 60º
Angle FBC = 30º as sin FBC = CF/BC = 1/2.
Now bisect angle ABF and you have the trisected 90º angle.

 

[soroban]

Hello, geekily!

Create an angle of 30º using only the figure below.

    A *
      |\
      | \
      |  \
      |40°\
      |    \
      |     \
      |      \
      |       \
      |        \
      |         \
      |       50°\
      * - - - - - *
      B           C

Let \(\displaystyle D\) be the midpoint of \(\displaystyle AC.\)
Draw \(\displaystyle DB\); extend \(\displaystyle BC\) to \(\displaystyle E.\;\angle DCE \,=\,130^o\)

    A *
      |\
      | \
      |  \
      |40°\
      |    \
      |     * D
      |    / \
      |   /   \ 
      |  /     \
      | /       \
      |/      50°\  130°
      * - - - - - * - - - - *
      B           C         E

Then \(\displaystyle \Delta DBC\) is isosceles: \(\displaystyle \angle DBC \,=\,50^o,\;\angle BDC \,=\,80^o\)
. . Hence: \(\displaystyle \:\angle ADB \,=\,100^o\)

    A *
      |\
      | \
      |  \
      |40°\
      |    \
      | 100°* D
      |    / \
      |   /80°\ 
      |  /     \
      | /       \
      |/50°   50°\  130°
      * - - - - - * - - - - *
      B           C         E

Then: \(\displaystyle \:\angle DCE\,-\,\angle ADB \;=\;130^o\,-\,100^o\;=\;30^o\)

 

[geekily]

Thank you so much, both of you. I went to school early to get help from the teacher so I figured it out, but it's always good to know another way to do it. Thanks!