Question

What is $\angle ABC$ if FB of line AB is ${45}^{\circ }$ and BB of line BC is ${280}^{\circ }?$

• A. ${320}^{\circ }$
• B. ${90}^{\circ }$
• C. ${235}^{\circ }$
• D. ${125}^{\circ }$

Answer 1

The correct option is D ${125}^{\circ }$

$BBofBC=FBofBC+{180}^{\circ }$
$\Rightarrow FBofBC={280}^{\circ }-{180}^{\circ }$
$\Rightarrow FBofBC={100}^{\circ }$

$BBofAB=FBofAB+{180}^{\circ }$
$={45}^{\circ }+{180}^{\circ }={225}^{\circ }$
$\therefore$ Included angle,
$\alpha =BBofAB-FBofBC$
$={225}^{\circ }-{100}^{\circ }={125}^{\circ }$
Alternatively,

$\therefore \angle ABC={125}^{\circ }$