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.


ANSWER

$$60^o$$


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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