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.

Answer :

Using the trigonometric ratio,

[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \text{where } \\ \theta=9^0 \\ \text{opposite}=h \\ \text{adjacent}=796ft \end{gathered}[/tex]

By substitution, we will have

[tex]\begin{gathered} \tan 9^0=\frac{h}{796ft} \\ h=796\times\tan 9^0 \\ h=796\times0.1584 \\ h=126.0864ft \end{gathered}[/tex]

Using trigonometric ratio also

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

By substitution, we will have

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

by equating the two values of h we will have

[tex]\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 \\ \frac{0.0699x}{0.0699}=\frac{70.446ft}{0.0699} \\ x=1007.8ft \\ to\text{ the nearest foot } \\ x=1008ft \end{gathered}[/tex]

Hence,

The distance of point A from point B = 1008ft

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