Question

In the figure $\overline{AB}$ = $\overline{AC}$, angle BAD = ${30}^{\circ }$ and $\overline{AE}$= $\overline{AD}$.Then x equals:

• A. $7{\frac{1}{2}}^{\circ }$
• B. ${10}^{\circ }$
• C. $12{\frac{1}{2}}^{\circ }$
• D. ${15}^{\circ }$
• E. ${20}^{\circ }$

Answer 1

Answer: (D)

${15}^{\circ }$

$AB=ACLet\angle B=\angle C=0\angle A=180°-20°\angle DAC=150°-20°\overline{AD}=\overline{AE}\angle ADE=\frac{180°-\angle DAC}{2}\angle ADE=15°+\theta$

$\angle ADC=\angle ABD+\angle DAB$ (Exterior angle = Sum of two internal opposite angles)

$15°+\theta +x=30°+\theta \therefore x=15°$