Question

$\text{What is}\angle ABC\text{if FB of line AB is}{40}^{\circ }\text{and BB of line BC is}{280}^{\circ }?$

• A. ${90}^{\circ }$
• B. ${120}^{\circ }$
• C. ${240}^{\circ }$
• D. ${320}^{\circ }$

Answer 1

The correct option is B ${120}^{\circ }$

$\text{BB of BC = FB of BC}+{180}^{\circ }$
$\Rightarrow \text{FB of BC}={280}^{\circ }-{180}^{\circ }$
$\Rightarrow \text{FB of BC}={100}^{\circ }$

$\text{BB of AB = FB of AB}+{180}^{\circ }$
$={40}^{\circ }+{180}^{\circ }={220}^{\circ }$
$\therefore \text{Included angle,}\alpha =\text{BB of AB - FB of BC}$
$={220}^{\circ }-{100}^{\circ }={120}^{\circ }$