Given: The information of a boat heading towards a lighthouse
To Determine: The distance from point A to point B
Solution: The information provided can be translated into the diagram below
[tex]\begin{gathered} m\angle ADC+m\angle DAC=90^0 \\ m\angle ADC+8^0=90^0 \\ m\angle ADC=90^0-8^0 \\ n\angle ADC=82^0 \end{gathered}[/tex][tex]\begin{gathered} m\angle BDC+m\angle DBC=90^0 \\ m\angle BDC=90^0-m\angle DBC \\ m\angle BDC=90^0-16^0 \\ m\angle BDC=74^0 \end{gathered}[/tex]Using SOH CAH TOA
[tex]\begin{gathered} \tan 74^0=\frac{BC}{108} \\ BC=108\tan 74^0 \\ BC=108\times3.4874 \\ BC=376.64ft \\ BC\approx377ft(nearest\text{ foot)} \end{gathered}[/tex][tex]\begin{gathered} \tan 82^0=\frac{AC}{108} \\ AC=108\times\tan 108 \\ AC=108\times7.115 \\ AC=768.4599 \end{gathered}[/tex][tex]\begin{gathered} AB=AC-BC \\ AB=768.45993-376.64 \\ AB=391.8199 \\ AB\approx392ft \end{gathered}[/tex]Hence, the distance from point A to point B is 392ft (nearest foot)
