Question

In the given figure, ABCDE is a pentagon inscribed in a circle. If AB = BC = CD, $\angle BCD={110}^o$ and $\angle BAE={120}^o$, find :
(i) $\angle ABC$ (ii) $\angle CDE$ (iii) $\angle AED$ (iv) $\angle EAD$

• A. (i) $\angle ABC={110}^o$
  (ii) $\angle CDE={95}^o$
  (iii) $\angle AED={105}^o$
  (iv) $\angle EAD={50}^o$
• B. (i) $\angle ABC={100}^o$
  (ii) $\angle CDE={95}^o$
  (iii) $\angle AED={105}^o$
  (iv) $\angle EAD={50}^o$
• C. (i) $\angle ABC={110}^o$
  (ii) $\angle CDE={95}^o$
  (iii) $\angle AED={125}^o$
  (iv) $\angle EAD={50}^o$
• D. (i) $\angle ABC={110}^o$
  (ii) $\angle CDE={95}^o$
  (iii) $\angle AED={105}^o$
  (iv) $\angle EAD={60}^o$

Answer 1

The correct option is A (i) $\angle ABC={110}^o$

(ii) $\angle CDE={95}^o$
(iii) $\angle AED={105}^o$
(iv) $\angle EAD={50}^o$
oin BD, BE
From isosceles $△BCD$ we get
$\angle CBD=\angle CDB=\frac{1}{2}({180}^{\circ }-{110}^{\circ })=\frac{1}{2}\times {70}^{\circ }={35}^{\circ }$
From the cyclic quadrilateral ABDE
$\angle BDE={180}^{\circ }-\angle BAE={180}^{\circ }-{120}^{\circ }={60}^{\circ }$
From the cyclic quadrilateral BCDE
$\angle BED={180}^{\circ }-{110}^{\circ }={70}^{\circ }$
$ArcAB=ArcBC=\angle AEB=\angle BDC={35}^{\circ }$
Sum of all the interior angles of a Pentagon $=(2\times 5-4)rt.\angle s=6\times {90}^{\circ }={540}^{\circ }$
(i) $\angle ABC={540}^{\circ }-({110}^{\circ }+{35}^{\circ }+{60}^{\circ }+{70}^{\circ }+{35}^{\circ }+{120}^{\circ })={540}^{\circ }-{430}^{\circ }={110}^{\circ }$
(ii) $\angle CDE=\angle CDB+\angle BDE={35}^{\circ }+{60}^{\circ }={95}^{\circ }$
(iii) $\angle AED=\angle AEB+\angle BED={35}^{\circ }+{70}^{\circ }={105}^{\circ }$
(iv) $\angle EAD=\angle EBD={180}^{\circ }-({70}^{\circ }+{60}^{\circ })={180}^{\circ }-{130}^{\circ }={50}^{\circ }$