Question

In the given figure, find the value of $\angle BAC$. $\Delta DEF$ is equilateral, $\angle CBA={40}^{\circ }$, $\angle FEC={50}^{\circ }$ and DE is parallel to BC. [4 MARKS]

Answer 1

Steps: 3 Marks
Result: 1 Mark

Given $DE\parallel BC,\angle CBA={40}^{\circ }$

$\Rightarrow \angle ADE=\angle CBA={40}^{\circ }$ [corresponding angle]

Since, $\Delta DEF$ is equilateral

$\Rightarrow \angle DEF={60}^{\circ }$

$\angle FEC={50}^{\circ }$ (Given)
$\angle AED={180}^{\circ }-({50}^{\circ }+{60}^{\circ })$ (Angles on a straight line)
= ${70}^{\circ }$
Consider $\Delta ADE$ ,

$\angle AED={70}^{\circ }$ & $\angle ADE={40}^{\circ }$

$\angle ADE+\angle AED+\angle DAE={180}^{\circ }$ [Angle sum property]

$\Rightarrow \angle DAE={180}^{\circ }-({70}^{\circ }+{40}^{\circ })={70}^{\circ }$.

$\therefore \angle BAC=\angle DAE={70}^{\circ }$.