All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Angle ABC (Posted on 2006-02-25)
In a triangle ABC, D is the midpoint of BC. Join AD. Angle ADB = 45 degree and angle ACB = 30 degree. Find angle ABC.

 No Solution Yet Submitted by akash Rating: 2.3000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Another approach | Comment 10 of 13 |
A similar approach to that of vije...

Extend CB to O, so that AO is perpendicular to CO.

WLOG, letting CD = DB = 1,
cot 30° = (2 + BO)/AO = sqrt(3),
cot 45° = (1 + BO)/AO = 1.
Hence BO/AO = cot ABO = 2 - sqrt(3).
(The three terms form an arithmetic progression.)

Now we can either note(!) that cot 75° = 2 - sqrt(3), or note that we can use the addition formula for cotangents:

cot 75° = cot (45° + 30°)
= (cot 45° cot 30° - 1)/(cot 45° + cot 30°)
= (sqrt(3) - 1)/(1 + sqrt(3))
= 2 - sqrt(3)

Either way, angle ABO = 75°, and so angle ABC = 105°.
Edited on February 26, 2006, 1:00 pm
 Posted by Nick Hobson on 2006-02-26 12:57:07
Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information