Question

In the given figure, $ABC$ is a triangle, side $CB$ is produced to $E$ and $\angle A$: $\angle B$: $\angle C$ $=2:1:3$. Find $\angle DBE$, if $DB$ is perpendicular to $AB$.

• A. ${66}^o$
• B. ${46}^o$
• C. ${60}^o$
• D. none of the above

Answer 1

Answer: (C)

${60}^o$

In triangle $ABC$, $\angle A:\angle B:\angle C=2:1:3$
Let the angles be $2x,x,3x$
By angle sum property, the sum of the angles $2x+x+3x={180}^o$
$⟹$ $6x={180}^o$
$⟹$ $x={30}^o$.

Hence, $\angle B={30}^o$.

Now,
$\angle DBE+\angle DBA+\angle ABC={180}^o$ ...[Straight angle property]
$\angle DBE+{90}^o+{30}^o={180}^o$
$\angle DBE={180}^o-{120}^o$
$\angle DBE={60}^{\circ }$.

Therefore, option $C$ is correct.