Hello, fireballs619!
Given: $\displaystyle PQ \parallel RS,\;\angle PAB = 115^o,\;\angle RCB = 135^o$
Find: $\displaystyle x \:=\:\angle ABC$ Code:
A
P * - - * - - - - - - - - * Q
115 *
*
*
*
*
x * B
*
*
*
*
135 *
R * - - - - - * - - - - * S
C
Extend $\displaystyle CB$ to intersect $\displaystyle PQ$ at $\displaystyle \,D.$
Code:
A D
P * - - * - - - - - - - - * - Q
115 * 65 45 /
* /
* /
* 70 /
* /
x * B
*
*
*
*
135 * 45
R * - - - - - * - - - - * S
C
$\displaystyle \angle PAB = 115^o \quad\Rightarrow\quad \angle QAB = 65^o$
$\displaystyle \angle RCB = 135^o \quad\Rightarrow\quad \angle BCS = 45^o = \angle ADB$
$\displaystyle \text{Hence: }\:\angle ABD \:=\:180^o - 65^o - 45^o \;=\;70^o$
$\displaystyle \text{Therefore: }\:x \;=\;\angle ABC \;=\;180^o - 70^o \;=\;110^o$