Little problem to keep sharp

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October 21st 2010, 05:49 PM
fireballs619

1 Attachment(s)

Little problem to keep sharp

Came across this problem today, and thought it was pretty nice. Most of the kids in my class solved it in about 1-2 minutes, how quickly can you solve it?

Attachment 19411

I drew it in paint, so pardon sloppiness (I don't want to mess with latex today).

Given:
m||n;
All lines are coplanar;

Find x

Spoiler:

x=110
October 21st 2010, 06:09 PM
Also sprach Zarathustra

I think that something is missing... are the two points make a perpendicular to m (n) ? is it matter? (asking myself)
October 21st 2010, 06:37 PM
skeeter

nothing missing ... sketch an auxiliary line parallel to m and n thru the vertex of angle x
October 22nd 2010, 08:24 AM
Soroban

Hello, fireballs619!

Quote:

Given: $PQ \parallel RS,\;\angle PAB = 115^o,\;\angle RCB = 135^o$

Find: $x \:=\:\angle ABC$

Code:

A P * - - * - - - - - - - - * Q 115 * * * * * x * B * * * * 135 * R * - - - - - * - - - - * S C

Extend $CB$ to intersect $PQ$ at $\,D.$

Code:

A D P * - - * - - - - - - - - * - Q 115 * 65 45 / * / * / * 70 / * / x * B * * * * 135 * 45 R * - - - - - * - - - - * S C

$\angle PAB = 115^o \quad\Rightarrow\quad \angle QAB = 65^o$

$\angle RCB = 135^o \quad\Rightarrow\quad \angle BCS = 45^o = \angle ADB$

$\text{Hence: }\:\angle ABD \:=\:180^o - 65^o - 45^o \;=\;70^o$

$\text{Therefore: }\:x \;=\;\angle ABC \;=\;180^o - 70^o \;=\;110^o$
October 22nd 2010, 08:36 AM
Unknown008

I just make use of the parallel lines and the time to grab my calculator, it takes less than 20 seconds.

http://p1cture.me/images/13385464636468101734.png

So, $x = \alpha + \beta = (180 - 115) + (180 - 135) = 110^o$