Question

In the given figure, points A, O and B are collinear.
Find the value of $\angle BOC$.

• A. ${60}^{\circ }$
• B. ${180}^{\circ }-{140}^{\circ }$
• C. ${40}^{\circ }$
• D. ${180}^{\circ }-{160}^{\circ }$

Answer 1

Answer: (B), (C)

The correct options are
B ${180}^{\circ }-{140}^{\circ }$
C ${40}^{\circ }$

Since A, O and B are collinear,
i.e., $\angle AOB={180}^{\circ }$
or, $\angle AOD+\angle DOC+\angle BOC={180}^{\circ }$
$(x-10{)}^{\circ }+(4x-25{)}^{\circ }+(x+5{)}^{\circ }={180}^{\circ }$
Or, $x+4x+x-{10}^{\circ }-{25}^{\circ }+5^{\circ }={180}^{\circ }$
Or, $6x-{30}^{\circ }={180}^{\circ }$
Or, $6x={180}^{\circ }+{30}^{\circ }={210}^{\circ }$
Or, $x=\frac{{210}^{\circ }}{6}={35}^{\circ }$
$\angle BOC=x+5^{\circ }={35}^{\circ }+5^{\circ }={40}^{\circ }$