Solution
Lighthouse B is 8 miles from lighthouse A,
The diagram below shows the representation of the details of the question
Let x represents the distance of the boat from B
To find x, we use the cosine rule which is
[tex]b^2=a^2+c^2-2ac\cos B_{}_{}[/tex]
Where
[tex]\begin{gathered} a=x \\ b=6 \\ c=8 \\ B=25\degree \end{gathered}[/tex]
Substitute the values into the formula above
[tex]\begin{gathered} 6^2=x^2+8^2-2(x)(8)\cos 25\degree \\ 36=x^2+64-16x\cos 25\degree \\ x^2+64-36-16x\cos 25\degree=0 \\ x^2+28-14.5x=0 \\ x^2-14.5x+28=0 \end{gathered}[/tex]
Solving for x, using the quadratic formula
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Where
[tex]a=1,b=-14.5,c=28[/tex]
Substitute the values of a, b and c into the formula above
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-(-14.5)\pm\sqrt[]{(-14,5)^2-4(1)(28)}_{}}{2(1)} \\ x=\frac{14.5\pm\sqrt[]{210.25-112}}{2} \\ x=\frac{14.5\pm\sqrt[]{98.25}}{2} \end{gathered}[/tex]
The values of x will be
[tex]\begin{gathered} x=\frac{14.5\pm\sqrt[]{98.25}}{2} \\ x=\frac{14.5\pm9.9121}{2} \\ x=\frac{14+9.9121}{2}=12.2\text{ (nearest tenth)} \\ x=\frac{14-9.9121}{2}=2.3\text{ (nearest tenth)} \end{gathered}[/tex]
The values of x are
[tex]x=12.2\text{ or 2.3}[/tex]
Hence, the boat is either 12.2 miles or 2.3 miles (nearest tenth) from lighthouse B.
