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.
Then the value of $\angle$DBE =${60}^{\circ }$, if DB is perpendicular to AB.( Enter 1 if true or 0 otherwise)

Answer 1

In triangle $ABC$,
$\angle A:\angle B:\angle C=2:1:3$
Let the angles be $2x,x,3x$
Sum of the angles of a triangle $=2x+x+3x={180}^o$
$6x={180}^o$
$x={30}^o$
Thus, $\angle ABC={30}^o$
Now, $\angle DBE+\angle DBA+\angle ABC={180}^o$ (Sum of angles of triangle)
$\therefore \angle DBE+{90}^o+{30}^o={180}^o$
$\therefore \angle DBE={60}^{\circ }$