Chapter 6.PS, Problem 1PS

Distance. In the figure, a beam of light is directed at the blue mirror, reflected to the red mirror, and then reflected back to the blue mirror. Find PT, the distance that the light travels from the red mirror back to the blue mirror.

To determine

To find:

The distance PT traveled by the light from the red mirror back to the blue mirror if the beam of light is directed at blue mirror, reflected to the red mirror, and then reflected back to the blue mirror.

Answer to Problem 1PS

Solution:

The distance is $PT=2.01\text{ feet}$.

Explanation of Solution

Formula:

Law of Cosines:

The formula for Law of Cosines is given by

$a^2=b^2+c^2-2bc\cos A$

Law of Sines:

The formula for Law of Sines is given by

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Calculation:

Consider a beam of light which is directed at the blue mirror, reflected to the red mirror and then reflected back to the blue mirror as shown in the figure.

The required distance is PT. The distance PT can be calculated only if the length of the either of the sides of the triangle OPQ. So, use Cosine Law to calculate the side PQ. So,

$\begin{array}{rll}{\left(\overline{PQ}\right)}^2& =& {\left(4.7\right)}^2+{\left(6\right)}^2-2\left(4.7\right)\left(6\right)\cos {25}^{\circ }\\ & =& 22.09+36-51.12\\ & =& 6.97\end{array}$

Take square root. So,

$\overline{PQ}\approx 2.64\text{ feet}$

Use the same triangle OPQ and use Sine Law to calculate the angle $\alpha$. So,

$\begin{array}{rll}\frac{4.7}{\sin \alpha }& =& \frac{2.64}{\sin {25}^{\circ }}\\ \sin \alpha & =& \frac{\left(4.7\right)\left(0.4226\right)}{2.64}\\ & =& 0.7523\end{array}$

Take arcsine on both the sides of the equation.

$\begin{array}{l}\alpha =\arcsin \left(0.7523\right)\\ \alpha ={48.79}^{\circ }\end{array}$

Denote the angle TPQ by $\beta$. In triangle OPQ, consider that sum of the angles in the triangle is ${180}^{\circ }$. So,

$\begin{array}{rll}{25}^{\circ }+\alpha +\theta +\beta & =& {180}^{\circ }\\ {25}^{\circ }+{48.79}^{\circ }+\theta +\beta & =& {180}^{\circ }\\ \theta +\beta & =& {180}^{\circ }-{25}^{\circ }-{48.79}^{\circ }\\ \theta +\beta & =& {106.21}^{\circ }\end{array}$

Use the fact that some of the angles in a straight line is ${180}^{\circ }$ for the red mirror. Hence,

$\theta +\beta +\theta ={180}^{\circ }$

It is obtained that $\theta +\beta ={106.21}^{\circ }$. So, substitute this value in the above equation to calculate $\theta$.

$\begin{array}{rll}\theta +\beta +\theta & =& {180}^{\circ }\\ {106.21}^{\circ }+\theta & =& {180}^{\circ }\\ \theta & =& {180}^{\circ }-{106.21}^{\circ }\\ \theta & =& {73.79}^{\circ }\end{array}$

Use this value in $\theta +\beta ={106.21}^{\circ }$ to calculate $\beta$.

$\begin{array}{rll}{73.79}^{\circ }+\beta & =& {106.21}^{\circ }\\ \beta & =& {106.21}^{\circ }-{73.79}^{\circ }\\ \beta & =& {32.42}^{\circ }\end{array}$

Denote the angle PTQ by $\gamma$ in triangle PTQ. Again use the fact that toms of the angles in the triangle PTQ is ${180}^{\circ }$. So,

$\begin{array}{rll}\alpha +\beta +\gamma & =& {180}^{\circ }\\ {48.79}^{\circ }+{32.42}^{\circ }+\gamma & =& {180}^{\circ }\\ \gamma & =& 180-{48.79}^{\circ }-{32.42}^{\circ }\\ \gamma & =& {98.79}^{\circ }\end{array}$

Now, denote the angle PTO by $\varphi$ in triangle OPT. Use the fact that some of the angles in a straight line is ${180}^{\circ }$ for the blue mirror. Hence,

$\begin{array}{c}\gamma +\varphi ={180}^{\circ }\\ {98.79}^{\circ }+\varphi ={180}^{\circ }\\ \varphi ={180}^{\circ }-{98.79}^{\circ }\\ \varphi ={81.21}^{\circ }\end{array}$

Use the Law of Sines for the triangle OPT to calculate PT. So,

$\begin{array}{rll}\frac{\overline{PT}}{\sin {25}^{\circ }}& =& \frac{4.7}{\sin {81.21}^{\circ }}\\ \overline{PT}& =& \frac{\left(4.7\right)\left(0.4226\right)}{0.9883}\\ & =& 0.7523\\ & \approx & 2.01\text{ feet}\end{array}$

Thus, the distance traveled by the light is $2.01\text{ feet}$.

Final Statement:

The light travels a distance of $2.01\text{ feet}$ from the red mirror back to the blue mirror.