Jump to content
BrainDen.com - Brain Teasers
  • 0

Question

Triangle ABC is isosceles.

Find angle x.

post-1048-1221180554_thumbgif

Do not assume the figure is drawn to scale.

Do not use the law of sines.

Use simple geometric reasoning, like


  • Isosceles triangles have two equal angles and sides.
  • The angles of a triangle sum to 180 degrees.
  • Things like that.

Have fun.

You can add a line or two if you think it will help.

Link to post
Share on other sites
  • Answers 66
  • Created
  • Last Reply

Top Posters For This Question

Recommended Posts

  • 0
Just to stop the pain in my head, I plugged this geometry into AutoCAD, and the answer is definitely 20 degrees. But for the life of me, I can't find the answer without using sine/cosine or some other method.

but that is just it

you cant find it without using those.

edit:

might i mention thank you for giving me the best/worst migraine of my life :lol:

Edited by solemnraven
Link to post
Share on other sites
  • 0

Tossing a couple Excedrin to SR. ;)

Top to bottom:

Angle at C: 20o

Angles at E: 150o-x, x, 30o

Angles at D: x+10o, 130o-x, 40o

Angles at crossing point: 50o, 130o, 130o, 50o

Angles at A; 10o, 70o

Angles at B: 20o, 60o

You can use triangle congruency [sAS] and [ASA] and the like.

Just keep it geometrical - no sine and cosine formulas.

And you can read the OP spoiler.

Link to post
Share on other sites
  • 0

Draw the red lines to create points L, M, N:

post-1048-1221472508_thumbgif

[1] Angle CAB = Angle CBA [triangle ABC is isosceles] = (180-20)/2 = 80

[2] Angle ACL [20/2] = Angle CAL [80-60-10] = 10. So triangle ACL is isosceles.

[3] AL = CL

[4] Angle LAM = Angle ECL = 10

[5] Angle ALM = Angle CLE [opposite angles of intersecting line segements]

[6] Triangles ALM and CEL have two angles[4,5] and included side[3] equal thus are congruent [Mirror Image].

[7] EL = LM [corresponding sides of congruent triangles.]

[8] CM = CL+LM and AE = AL+LE [inspection]

[9] CM = AE [3,7,8]

[10] Angle ACN (20) = Angle CAN [equal to CBD(20) by symmetry]. So triangle ACN is isosceles.

[11] AN = CN

[12] Angle EAN [80-60-10] = Angle MCN [20/2].

[13] Triangles AEN and CMN have two sides[9,11] and included angle[12] equal thus are congruent [Mirror Image].

[14] EN = MN [corresponding sides of congruent triangles.]

Angle BAN = Angle ABD(60) [symmetry] so triangle ABM is equilateral.

DN is parallel to AB [AN is symmetrical to BD] so Triangle DMN is equlateral.

[15] Angle DNE = Angle ABE = 80

[16] DM = DN = MN

DN = EN [14, 16] so Triangle DEN is isosceles.

[17] Angle EDN = Angle DEN = (180 - DNE)/2 = (180-80[16])/2 = 50

[18] Angle AEB = 180 - (Angle EAB + Angle ABE) = 180 - (70[given] + 80[15]) = 30

So finally,

x = Angle DEN - Angle AEB [inspection] = 20 [17, 18].

Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
  • Recently Browsing   0 members

    No registered users viewing this page.


×
×
  • Create New...