Question

Mr. Miller's field of vision is 140 degrees, as shown in the diagram below. From his beach house he can see ships on the horizon up to 4 miles away.To the nearest tenth of a mile, how many miles of the horizon can Mr. Miller see along the arc of his field of vision?A. 9.8 milesB. 12.6 milesC. 25.1 milesD. 19.5 miles

Answer 1

`\begin{gathered} the\text{ relations betwe}en\text{ the arc-lenght (horizon), radius and angle} \\ is\text{ given by} \\ \theta=(s)/(r) \\ \text{where} \\ \theta\text{ is the angle in radians} \\ s\text{ is the arc-lenght (horizon)} \\ r\text{ is the radius} \\ In\text{ this case, we have:} \end{gathered}`
`\begin{gathered} \theta=140\text{ degre}es \\ r=4\text{ miles} \\ \text{from the above relationship, we have} \\ s=\theta\cdot r \\ \text{however, in order to use this formula, we must convert} \\ 140\text{ degre}es\text{ to radians.} \\ \text{this can be obtained} \end{gathered}`
`\begin{gathered} by\text{ using the rule of thre}e\colon \\ 1\pi\text{ radian ----180 degre}es \\ x\text{ --------140 degre}es \\ \text{then,} \\ x=((\pi)(140))/(180) \\ x=((\pi)(7))/(9) \\ x=2.44\text{ radian} \\ \text{therefore, 144 degre}es\text{ is equivalent to 2.44 radians} \\ i.e\text{. }\theta=2.44\text{ radians.} \end{gathered}`
`\begin{gathered} \text{Now, we can use the above formula:} \\ s=\theta\cdot r \\ s=(2.44)(4)\text{ miles} \\ s=9.77\text{ miles} \\ \text{hence, Mr Miller's field of vision is 9.8 miles} \end{gathered}`