Subject: Mathematics — Trigonometry
Question: A triangle $ABC$ has sides $AB = AC = 5\ \text{cm}$ and
$BC = 6\ \text{cm}$. Triangle $A'B'C'$ is the reflection of the triangle
$ABC$ in a line parallel to $AB$ placed at a distance $2\ \text{cm}$
from $AB$, outside the triangle $ABC$. Triangle $A''B''C''$ is the
reflection of the triangle $A'B'C'$ in a line parallel to $B'C'$ placed
at a distance of $2\ \text{cm}$ from $B'C'$ outside the triangle
$A'B'C'$. Find the distance between $A$ and $A''$.
Answer: $8\sqrt{\frac{17}{5}}$
Solution: Let $L$ be the line parallel to side $AB$ of
$\bigtriangleup ABC$, at a distance of $2\ \text{cm}$ from $AB$, in
which the first reflection $\bigtriangleup A'B'C'$ is obtained. Let $L'$
be the second line parallel to $B'C'$, at a distance of $2\ \text{cm}$
from $B'C'$, in which reflection of $\bigtriangleup A'B'C'$ is taken as
$\bigtriangleup A''B''C''$.
In figure, size of $\bigtriangleup A''B''C''$ is same to the size of
$\bigtriangleup A'B'C'$.
In the figure, distance between $AB$ & $A'B'$ and distance between
$B'C'$ & $B''C''$ will be same.
[IMAGE] From figure $AA' = 4\ \text{cm}$ and $A'A'' = 12\ \text{cm}$. So to
find $AA''$ it suffices to know $\angle AA'A''$, clearly
$$\angle AA'A'' = 90^{\circ} + \alpha,\quad\text{where}\quad\sin\alpha = \frac{3}{5}.$$
$$\Rightarrow cos\angle AA'A'' = cos\left( 90^{\circ} + \alpha \right) = - sin\alpha = - \frac{3}{5}.$$
Now, in $\bigtriangleup AA'A''$,
$$\cos\left( 90^{\circ} + \alpha \right) = \frac{(AA')^{2} + (A'A'')^{2} - (AA'')^{2}}{2(AA')(A'A'')}\quad\quad\left\lbrack \text{cosine law} \right\rbrack$$
$$\therefore AA'' = \sqrt{(AA')^{2} + (A'A'')^{2} - 2AA' \times A'A'' \cdot cos\left( 90^{\circ} + \alpha \right)}$$
$$= \sqrt{16 + 144 + 96 \times \frac{3}{5}} = \sqrt{\frac{1088}{5}} = 8\sqrt{\frac{17}{5}}\ \text{cm}.$$