Chapter 14.4, Problem 18PPSE

Solution

To find: The distance of fire from two opposite towers.

The distance of fire from first tower is 7.52 miles and from second tower 7.87 miles.

Given data: A ranger see fire from tower ${\text{T}}_{\text{1}}$ at angle ${39}^{\text{ο}}$ east of north and another ranger 10 miles from first tower see fire from tower ${\text{T}}_2$ at angle ${42}^{\text{ο}}$ west of north.

Method/Formula used:

In a triangle $\Delta ABC$ shown in Fig. 1 the sin rule states that

  

  $\begin{array}{c}\frac{1}{2}ab\sin C=\frac{1}{2}bc\sin A=\frac{1}{2}ca\sin B\\ \frac{\sin C}{c}=\frac{\sin A}{a}=\frac{\sin B}{b}\end{array}$

Calculations:

The position of first tower ${\text{T}}_{\text{1}}$ and second tower ${\text{T}}_2$ , 10 miles apart and the location of fire F is shown in Fig. 1.

  

Let the distance of fire F from tower ${\text{T}}_{\text{1}}$ is ‘ a’ and from tower ${\text{T}}_2$ is ‘ b’ . Fire, and two towers form a triangle $\Delta {\text{T}}_{\text{1}}\text{F}{\text{T}}_2$ in which

  $\begin{array}{l}\angle F{T}_{\text{1}}{T}_{\text{2}}={90}^{\text{ο}}-{39}^{\text{ο}}={51}^{\text{ο}}\\ \angle F{T}_{\text{2}}{T}_{\text{1}}={90}^{\text{ο}}-{42}^{\text{ο}}={48}^{\text{ο}}\\ \angle T_{\text{2}}F{T}_{\text{1}}={51}^{\text{ο}}+{48}^{\text{ο}}={99}^{\text{ο}}\end{array}$

For triangle $\Delta {\text{T}}_{\text{1}}\text{F}{\text{T}}_2$ , the sine rule states that,

  $\frac{\sin \angle F{T}_1{T}_2}{b}=\frac{\sin \angle F{T}_2{T}_1}{a}=\frac{\sin \angle T_{1}F{T}_2}{T_1{T}_2}$ … (1)

Substitute ${51}^{\text{ο}}$ for $\angle F{T}_{\text{1}}{T}_{\text{2}}$ , ${48}^{\text{ο}}$ for $\angle F{T}_{\text{2}}{T}_{\text{1}}$ , ${99}^{\text{ο}}$ for $\angle T_{\text{2}}F{T}_{\text{1}}$ , and 10 for $T_1{T}_2$ in result (1),

  $\frac{\sin {51}^{\text{ο}}}{b}=\frac{\sin {48}^{\text{ο}}}{a}=\frac{\sin {99}^{\text{ο}}}{10}$ … (2)

From first and third ratio of result (2),

  $\begin{array}{c}\frac{\sin {51}^{\text{ο}}}{b}=\frac{\sin {99}^{\text{ο}}}{10}\\ \Rightarrow b=\frac{10\sin {51}^{\text{ο}}}{\sin {99}^{\text{ο}}}\\ \Rightarrow b=7.87\end{array}$

From second and third ratio of result (2),

  $\begin{array}{c}\frac{\sin {48}^{\text{ο}}}{a}=\frac{\sin {99}^{\text{ο}}}{10}\\ \Rightarrow a=\frac{10\sin {48}^{\text{ο}}}{\sin {99}^{\text{ο}}}\\ \Rightarrow a=7.52\end{array}$

Therefore, the distance of fire from first tower is 7.52 miles and from second tower 7.87 miles.