Question

In the figure(not drawn to scale), $EFA$ is a right-angled triangle with $\angle$ $EFA$ $={90}^o$ and $FGB$ is an equilateral triangle, find $y-2x$.

• A. $2$
• B. $8$
• C. $17$
• D. $20$

Answer 1

The correct option is A $2$

Given that $FGB$ is an equilateral triangle

$\Rightarrow \angle BGF=\angle BFG={60}^{\circ }$

In a $\Delta FCG$, we know

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

$\Rightarrow \angle x+{92}^{\circ }+\angle FGC={180}^{\circ }$

$\Rightarrow \angle x+{92}^{\circ }+\angle FGB={180}^{\circ }$

$\Rightarrow \angle x+{92}^{\circ }+{60}^{\circ }={180}^{\circ }$

$\Rightarrow \angle x={28}^{\circ }$

$\angle BFG=\angle BFC+\angle x={60}^{\circ }$

$\Rightarrow \angle BFC={60}^{\circ }-{28}^{\circ }$

$\Rightarrow \angle BFC={32}^{\circ }$

$EFA$ is a right-angled triangle.

Thus $\angle EFA={90}^{\circ }$

We have $\angle EFA=\angle BFC+\angle y={90}^{\circ }$

$\Rightarrow {32}^{\circ }+\angle y={90}^{\circ }$

$\Rightarrow \angle y={58}^{\circ }$

We have to find:

$y-2x=58-2\times 28=58-56$

$\Rightarrow y-2x=2$