Question

The diagram shows $2$ isoceles triangles. What is the sum of $a$ and $b$.

• A. ${116}^{\circ }$
• B. ${79}^{\circ }$
• C. ${58}^{\circ }$
• D. ${137}^{\circ }$

Answer 1

Answer: (D)

The correct option is C ${137}^{\circ }$
In $\Delta ABC,CA=CB\Rightarrow \angle CBA=\angle CAB$ ..... $\left(Anglesoppositetoequalsidesareequal\right)$
$\therefore \angle CBA=\angle CAB=\frac{1}{2}({180}^{\circ }-\angle BCA)$
$=\frac{1}{2}({180}^{\circ }-{64}^{\circ })=\frac{1}{2}\times {116}^{\circ }={58}^{\circ }$
$\Rightarrow a={58}^{\circ }$
Now $\angle CAB=\angle CAE+\angle EAD+\angle DAB$
$\Rightarrow {58}^{\circ }={25}^{\circ }+\angle EAD+{11}^{\circ }$
$\Rightarrow \angle EAD={58}^{\circ }-{36}^{\circ }={22}^{\circ }$
$\therefore$ In $\Delta ADE,AD=AE$
$\Rightarrow \angle AED=\angle ADE=\frac{1}{2}({180}^{\circ }-{22}^{\circ })$
$=\frac{1}{2}\times {158}^{\circ }={79}^{\circ }$
$\Rightarrow b={79}^{\circ }$
Hence, $a+b={58}^{\circ }+{79}^{\circ }={137}^{\circ }$.