SSC CGL 10 Jun 2019 Shift 1 Paper

Question : 73 of 100

Marks: +1, -0

Let O be the centre of a circle and AC be its diameter. BD is a chord intersecting AC at E. Point A is joined toB and D. If ∠BOC = 50° and ∠AOD = 110°, then ∠BEC = ?

80°
70°
55°
90°

Validate

Solution:

\∠\ {\ABD}=\ {1}/{2} ∠{\AOD}=\ {110}/{2}=55°

Now,

\∠A B C=90

(angle on diameter)
So,

\∠E B C=90-55=35°

\ {\Now} \∠\ {\AOB}=180-50°

=130°

\ {\In} \Δ\ {\AOB}

\∠\ {\OAB}+\∠ \ {\ABO}+130=180

\∠ \ {\ABO}=\ {50}/{2}=25°(\ {\as} \∠\ {\OAB}=\∠\ {\ABO})

Since

\∠\ {\ABC}=90

\∠\ {\ABO}+\∠\ {\OBE}+\∠\ {\EBC}=90

25+\∠O B E+35=90°

\∠\ {\OBE}=30°

Now,

\∠\ {\EOB}+\∠\ {\OBE}=\∠\ {\BEC}

\⇒\∠B E C=50+30=80°

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