Question

If CE is parallel to DB in the given figure and ∠ADC = 70°, then the value of ‘x’ will be:

• A. 45°
• B. 50°
• C. 30°
• D. 60°

Answer 1

The correct option is A 45°



In △ABD:

$\angle ADB+{55}^{\circ }+{100}^{\circ }={180}^{\circ }$ [Angle sum property of a triangle]

$\Rightarrow \angle ADB+{155}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ADB={180}^{\circ }-{155}^{\circ }={25}^{\circ }$

Now, ∠ADB + ∠BDC = 70°

$\Rightarrow \angle BDC={70}^{\circ }-{25}^{\circ }={45}^{\circ }$

Now,
$\angle ECF=\angle BDC={45}^{\circ }$

(∵ Given:CE || DB, using corresponding angles axiom)

Now, DF is a straight line.
We know, adjacent angles on a straight line makes up 180°.

$\text{So,}\angle BCD+\angle BCE+\angle ECF={180}^{\circ }$

$\Rightarrow {50}^{\circ }+\angle BCE+{45}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BCE={180}^{\circ }-{95}^{\circ }$

$\Rightarrow \angle BCE={85}^{\circ }$

Now, BD and EC are parallel lines and ∠DCB and ∠CBE are interior alternate angles.

So, ∠DCB = ∠CBE = 50°

Angle sum property of a triangle tells sum of interior angles of a triangle sums upto 180°.

In △BCE:

∠CBE + ∠BCE + ∠BEC = 180°
⇒ 50° + 85° + x° = 180°
⇒ 135° + x° = 180°
⇒ x° = 180° - 135° = 45°