Question

In the given figure $AB\parallel CD$ and $EF\parallel GH$. If $\angle AIF={120}^o$, then $\angle CJG$ is

• A. ${80}^o$
• B. ${100}^o$
• C. $\frac{1}{3}$ of a right angle
• D. $\frac{2}{3}$ of a right angle

Answer 1

Answer: (D)

$\frac{2}{3}$ of a right angle

$\angle CLI={180}^o-\angle AIL={180}^o-{120}^o={60}^o$

($AB\parallel CD$ interior angles)

Now,

$\angle CJG=\angle CLI={60}^o=\frac{2}{3}\times {90}^o$

($EF\parallel GH$, corresponding angles)