GATE CS 2018 Q.9 (GA Q.9): Quantitative Aptitude

In the figure below, $\angle DEC + \angle BFC$ is equal to ____________ .

Geometric diagram showing points A, B, C, D, E, F with line segments EDA and ABF, and lines EB and FD intersecting at C, illustrating angles DEC and BFC.

$\angle BCD - \angle BAD$

$\angle BAD + \angle BCF$

$\angle BAD + \angle BCD$

$\angle CBA + \angle ADC$

Official answer and solution

Official answer: A
From: Official GATE 2018 answer key

Solution

Use the angle sum property of triangles $EAB$ and $FAB$ along with quadrilateral $ABCD$ .

In $\triangle EAB$ , points $E, D, A$ form a straight line and $E, C, B$ form a straight line.
The angle at $E$ is $\angle DEC$ , the angle at $A$ is $\angle BAD$ , and the angle at $B$ is $\angle ABE$ .
Sum of angles in $\triangle EAB$ is $180^\circ$ :

\angle DEC = 180^\circ - \angle BAD - \angle ABE

In $\triangle FAB$ , points $A, B, F$ form a straight line and $F, C, D$ form a straight line.
The angle at $F$ is $\angle BFC$ , the angle at $A$ is $\angle BAD$ , and the angle at $D$ is $\angle ADF$ .
Sum of angles in $\triangle FAB$ is $180^\circ$ :

\angle BFC = 180^\circ - \angle BAD - \angle ADF

Adding these two equations:

\angle DEC + \angle BFC = 360^\circ - 2\angle BAD - (\angle ABE + \angle ADF)

Now consider the interior angles of quadrilateral $ABCD$ at vertices $A, B, C, D$ :

\angle BAD + \angle ABE + \angle BCD + \angle ADF = 360^\circ

Therefore:

\angle ABE + \angle ADF = 360^\circ - \angle BAD - \angle BCD

Substitute this back into the sum:

\angle DEC + \angle BFC = 360^\circ - 2\angle BAD - (360^\circ - \angle BAD - \angle BCD) = \angle BCD - \angle BAD

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2018 Computer Science and Information Technology (CS)
Question: 9 of 65, printed as GA Q.9
Section: General Aptitude

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