Question

In the given figure, $AB=BC=CD=EG=GA$ and $\angle CAD={25}^{\circ }$. Then

• A. $\angle BCD={80}^{\circ }$
• B. $\angle CDA={60}^{\circ }$
• C. $\angle GEA={50}^{\circ }$
• D. $\angle ACD={125}^{\circ }$

Answer 1

The correct option is A $\angle BCD={80}^{\circ }$

In $△ABC$,
$AB=AC$
Hence, $\angle BAC=\angle BCA={25}^{\circ }$ (isosceles triangle property)
Sum of angles $=180$
$\angle BAC+\angle BCA+\angle ABC=180$
$25+25+\angle ABC=180$
$\angle ABC={130}^{\circ }$
Now, $\angle CBD=180-\angle ABC$
$\angle CB=180-130={50}^{\circ }$
In $△CBD$,
$BC=CD$
$\angle CBD=\angle CDB$ (Isosceles triangle property)
Sum of angles $=180$
$\angle CBD+\angle CDB+\angle BCD=180$
$\angle BCD=180-50-50$
$\angle BCD={80}^{\circ }$