Question

In the given regular polygon, find the value of $\angle$ ECB.

• A. ${54}^{\circ }$
• B. ${72}^{\circ }$
• C. ${36}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B

${72}^{\circ }$

In a regular pentagon all sides are equal in length and every angle measures ${108}^{\circ }$.
In $\Delta$ ABE and $\Delta$ EDC
AB = CD
AE = DE
$\angle$ BAE = $\angle$ CDE
$\Delta$ BAE $\cong$ $\Delta$ CDE
BE = EC (Since triangles are congruent)
$\angle$ ABE = $\angle$ AEB (Angles opposite to equal sides are equal)
$=x^0$
$\angle$ AEB = $\angle$ DEC (Since triangles are congruent corresponding angles are equal)
$\angle$ EBC = $\angle$ ECB (Angles opposite to equal sides are equal)
Let $\angle$ ECB = $y^0$
x + y = ${108}^{\circ }$ ...............(i)
In $\Delta$ BEC
$\angle$ BEC + $\angle$ ECB + $\angle$ CBE = ${180}^{\circ }$ (Sum of angles in a triangle)
2y - 2x = ${72}^{\circ }$
y - x = ${36}^{\circ }$ ................(ii)
Adding (i) and (ii) we get
$y={72}^{\circ }$