# Little problem to keep sharp

• Oct 21st 2010, 05:49 PM
fireballs619
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
• Oct 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)
• Oct 21st 2010, 06:37 PM
skeeter
nothing missing ... sketch an auxiliary line parallel to m and n thru the vertex of angle x
• Oct 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

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

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

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

$\text{Therefore: }\:x \;=\;\angle ABC \;=\;180^o - 70^o \;=\;110^o$

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