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A boat heading out to sea starts out at Point A, at a horizontal distance of 796 feetfrom a lighthouse/the shore. From that point, the boat's crew measures the angle ofelevation to the lighthouse's beacon-light from that point to be 9º. At some later time,the crew measures the angle of elevation from point B to be 4°. Find the distancefrom point A to point B. Round your answer to the nearest foot if necessary.

User Memento
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1 Answer

22 votes
22 votes

Using the trigonometric ratio,


\begin{gathered} \tan \theta=(opposite)/(adjacent) \\ \text{where } \\ \theta=9^0 \\ \text{opposite}=h \\ \text{adjacent}=796ft \end{gathered}

By substitution, we will have


\begin{gathered} \tan 9^0=(h)/(796ft) \\ h=796*\tan 9^0 \\ h=796*0.1584 \\ h=126.0864ft \end{gathered}

Using trigonometric ratio also


\begin{gathered} \tan \theta=(opposite)/(adjacent) \\ \text{where,} \\ \text{opposite}=h \\ \text{adjacent}=(796+x)ft \\ \theta=4^0 \end{gathered}

By substitution, we will have


\begin{gathered} \tan 4^0=(h)/((796+x)ft) \\ h=\text{tan}4(796+x) \\ h=0.0699(796+x) \\ h=55.6404+0.0699x \end{gathered}

by equating the two values of h we will have


\begin{gathered} 126.0864ft=55.6404ft+0.0699x \\ \text{collecting like terms we will have} \\ 0.0699x=126.0864-55.6404 \\ 0.0699x=70.446ft \\ (0.0699x)/(0.0699)=(70.446ft)/(0.0699) \\ x=1007.8ft \\ to\text{ the nearest foot } \\ x=1008ft \end{gathered}

Hence,

The distance of point A from point B = 1008ft

A boat heading out to sea starts out at Point A, at a horizontal distance of 796 feetfrom-example-1
User CLiown
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