Question

In the figure, $△EFA$ is a right-angled triangle with $\angle EFA={90}^{\circ }$ and $△FGB$ is an equilateral triangle, find $y–2x$.

• A. $2^{\circ }$
• B. $8^{\circ }$
• C. ${17}^{\circ }$
• D. ${20}^{\circ }$

Answer 1

The correct option is A $2^{\circ }$

Given,
$△FGB$ is an equilateral triangle.
$⟹\angle FGB={60}^{\circ }$

Now, In $△FGC$
$\angle FGC+\angle GCF+\angle CFG={180}^{\circ }\left[Anglesumproperty\right]⟹{60}^{\circ }+{92}^{\circ }+x={180}^{\circ }[\angle FGC=\angle FGB,\angle GCF={92}^{\circ }]⟹x={180}^{\circ }-{152}^{\circ }={28}^{\circ }$

And, $\angle BFG={60}^{\circ }\left[△FGBisequilateral\right]⟹\angle BFC+\angle CFG={60}^{\circ }⟹\angle BFC+x={60}^{\circ }⟹\angle BFC={60}^{\circ }-{28}^{\circ }={32}^{\circ }[x={28}^{\circ }]$

Now,
$\angle BFC+y={90}^{\circ }[\angle EFA={90}^{\circ }]⟹{32}^{\circ }+y={90}^{\circ }⟹y={90}^{\circ }-{32}^{\circ }={58}^{\circ }$

$⟹y-2x={58}^{\circ }-(2\times {28}^{\circ })={58}^{\circ }-{56}^{\circ }⟹y-2x=2^{\circ }$