Chapter 9.7, Problem 29E

a.

To determine

To find the value of distance $d\text{ when n = 4}\text{.}$

Answer to Problem 29E

Distance = $d=4400feet$ .

Explanation of Solution

Given:

  

Calculation:

  $\begin{array}{l}\text{Distance}=speed\ast time\\ \Rightarrow d=\text{ (1100 feet/seconds)}\ast (\text{4seconds)=4400 feet}\text{.}\\ \text{Distance = 4400 feet}\text{.}\end{array}$

b.

To determine

To find the value of $s$ .

Answer to Problem 29E

The value of $s=2552feet$ .

Explanation of Solution

Given:

  

Calculation:

  $\begin{array}{l}\Delta ABD=\Delta CBD={30}^0\\ For\text{ both }the\text{triangle }\Delta ABD,\\ \tan {30}^0=\frac{opposite}{adjacent}=\frac{s}{d}\text{ ----(1)}\\ \Rightarrow \text{0}\text{.58 = }\frac{s}{4400}\Rightarrow s=0.58\ast 4400=\text{ 2552 feet}\text{.}\\ \text{Hence the value s=2552 feet}\end{array}$

c.

To determine

To find the distance across the fireworks star to the nearest foot

Answer to Problem 29E

The distance across the fireworks star to the nearest foot $=765.6\text{}\text{foot}$

Explanation of Solution

Given: $\text{x=5}$

  

Calculation:

  $\begin{array}{l}Ifx=5\text{ , then tanx = }\frac{s}{d}\\ \Rightarrow {\text{tan5}}^0\text{ = }\frac{s}{4400}\Rightarrow s=0.087\ast 4400feet=382.8foot\\ Thenthe\text{ distance across the fireworks star to the nearest foot = s+s = }382.8\text{+}382.8\text{ = 765}\text{.6 foot}\end{array}$