Question

In figure AD=CD=BC and BEC =${96}^{\circ }$ then fing $\angle DBC$

• A. ${36}^{\circ }$
• B. ${32}^{\circ }$
• C. ${64}^{\circ }$
• D. ${72}^{\circ }$

Answer 1

Answer: (B)

${32}^{\circ }$

Given AD=$CD$=$BC,

$\therefore \angle$CAD$=\angle$DCA and $\angle$CDB$=\angle$CBD

By Exterior angle property $2x=y$ ……..$\left(1\right)$

in $\Delta$BCD, $180-2y=\angle$BCD

and also $96+180-2y+x=180$

$96+180-4x+x=180$

$\Rightarrow x=32$, substituting in $\left(1\right)$ $y=64=\angle$DBC.