Question

In the given figure, if $\angle ACE={43}^{\circ }$ and $\angle CAF={62}^{\circ }$, then match the following correctly.

$\begin{array}{cc}ListI& ListII\\ A.\angle a& i{.62}^{\circ }\\ B.\angle c& ii.{105}^{\circ }\\ C.\angle b& iii.{13}^{\circ }\end{array}$

• A. A-i; B-ii; C-iii
• B. A-iii; B-i; C-ii
• C. A-ii; B-i; C iii
• D. A-ii; B-iii; C-i

Answer 1

The correct option is C

A-ii; B-i; C iii

In $\Delta ACE$
$\angle ACE+\angle CEA+\angle CAE={180}^{\circ }\Rightarrow {43}^{\circ }+\angle CEA+{62}^{\circ }={180}^{\circ }\therefore \angle CEA={180}^{\circ }-{105}^{\circ }={75}^{\circ }$
In cyclic quadrilateral BDEA,
$\angle a+\angle CEA={180}^{\circ }\therefore \angle a={180}^{\circ }-{75}^{\circ }={105}^{\circ }$
In cyclic quadrilateral BDEA,
$\angle BAE+\angle BDE={180}^{\circ }\therefore \angle BAE+{180}^{\circ }-\angle c={180}^{\circ }(\because \angle c+\angle BDE={180}^{\circ })\Rightarrow \angle BAE=\angle c={62}^{\circ }$
In $\Delta BFA,$
$\angle a+\angle BAF+\angle b={180}^{\circ }\Rightarrow {105}^{\circ }+{62}^{\circ }+\angle b={180}^{\circ }\therefore \angle b={180}^{\circ }-{167}^{\circ }={13}^{\circ }\angle b={13}^{\circ }$