Question

In the given figure, $△ABC$ is an equilateral triangle and $\square AWXB$ and $\square AYZC$ are two squares. The value of $\frac{1}{10}\left(\angle ZXA\right)$ is:

Answer 1

Given that $ABC$ is an equilateral triangle.

Therefore, $\angle ABC=\angle CBA=\angle BAC={60}^o$

GIven that $ABXW$ and $AYZC$ are two squares.

Therefore, $AX$ is a diagonal of square $ABXW$ - as shown in the figure.

All interior angles of a square are equal to ${90}^o$

Therefore, $\angle BAW={90}^o$

The diagonal of a square bisects the angle at the vertex.

Therefore, $\angle BAX=\angle AXW={45}^o$

$\Rightarrow \angle OAX={45}^o$ -------$\left(1\right)$

Join vertices $X$ and $Z$ - as shown in the figure. Join $ZX$, a straight line.

Line $ZX$ cuts the sides $AB$ and $AC$, of the Equilateral triangle $ABC$, at $O$ and $N$ respectively.

The smaller triangle $AON$ is similar to triangle $ABC$ are similar, thus triangle $AON$ is also an equilateral triangle.

Therefore, $\angle NAO=\angle AON=\angle ANO={60}^o$

Line $A0$ is cutting the straight line $XZ$, hence

$\angle ZOA+\angle XOA=\angle ZOX={180}^o$

${60}^o+\angle XOA={180}^o$

$\angle XOA={180}^o-{60}^o$

$\angle XOA={120}^o$ -------$\left(2\right)$

Consider triangle $AOX$ - in figure:

Here, $\angle XOA+\angle OAX+\angle OXA={180}^o$

${120}^o+{45}^o+\angle OXA={180}^o$, from equations $\left(1\right)$ and $\left(2\right)$

$\angle OXA={180}^o-{120}^o-{45}^o$

$\angle OXA={15}^o$

$\angle OXA=\angle ZXA={15}^o$

Therefore,

$\frac{1}{10}\angle ZXA=\frac{1}{10}\times {15}^o$

$\frac{1}{10}\angle ZXA={1.5}^o$