Given figurc
We find thc volue of t
From thc figure we sec thc t $\mathrm{\angle}AOB+\mathrm{\angle}BOC=180\xb0$ $[\therefore \mathrm{\angle}AOB{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\mathrm{\angle}BOC\text{}are\text{}lincar\text{}pair]$ $\Rightarrow 41\xb0+t=180\xb0$ $[\therefore \mathrm{\angle}AOB=41\xb0\mathrm{\angle}BOC=t]$ $\Rightarrow t=180\xb0-41\xb0$ $\Rightarrow t=139\xb0$
Hence $t=139\xb0$