Question

In the figure (not drawn to scale), $ABC$ is an equilateral triangle and $ABD$ is an isosceles triangle with $DA$ $=$ $DB$, find $x$.

• A. ${14}^o$
• B. ${16}^o$
• C. ${12}^o$
• D. ${32}^o$

Answer 1

Answer: (A)

${14}^o$

Given that $\Delta ABC$ is an equilateral triangle

$\Rightarrow \angle A=\angle B=\angle C={60}^{\circ }$

In a $\Delta DAB$,

$DA=DB$ ....Given

$\Rightarrow \angle ABD=\angle DAB$ angles opposite to equal sides are also equal

Sum of all angles of a triangle $={180}^{\circ }$.

$\Rightarrow \angle ABD+\angle DAB+{88}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ABD+\angle DAB={92}^{\circ }$

$\Rightarrow \angle ABD=\angle DAB={46}^{\circ }$

$\Rightarrow \angle B=\angle x+\angle ABD={60}^{\circ }$

$\Rightarrow \angle x+{46}^{\circ }={60}^{\circ }$

$\Rightarrow \angle x={14}^{\circ }$