Mathematics

# 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.

$60^o$

##### SOLUTION
In triangle ABC, $\angle A : \angle B : \angle C = 2: 1: 3$
Let the angles be $2x, x, 3x$
Sum of the angles $2x + x+ 3x = 180$
$6x = 180$
$x = 30$
Hence, $\angle B = 30$
Now,
$\angle DBE + \angle DBA + \angle ABC = 180$
$\angle DBE + 90 + 30 = 180$
$\angle DBE = 180 - 120$
$\angle DBE = 60^{\circ}$

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Single Correct Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
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