Question

In the given figure, if $\angle BOC=7x+{20}^{0}and\angle COA=3x$, then the value of x for which AOB becomes a straight line is

• A. ${16}^0$
• B. ${14}^0$
• C. ${20}^0$
• D. ${21}^0$

Answer 1

Answer: (A)

${16}^0$

For $\angle AOB$ to be a straight line, it should be equal to ${180}^{\circ }$

$\Rightarrow 7x+{20}^{\circ }+3x={180}^{\circ }$

$\Rightarrow 10x={180}^{\circ }-{20}^{\circ }$

$\Rightarrow 10x={160}^{\circ }$

$\therefore x={16}^{\circ }$