AC is a straight shoreline and B is a boat out at sea. Find the shortest distance from the boat to the shore if A and C are 5 km apart, with angles A and C being 60° and 45° respec... AC is a straight shoreline and B is a boat out at sea. Find the shortest distance from the boat to the shore if A and C are 5 km apart, with angles A and C being 60° and 45° respectively. Read more

Steps to Solve

1. Draw a diagram and label the unknowns

Draw the triangle ABC and add the perpendicular line from B to AC, and call the point where it intersects D. Let BD = $h$, AD = $x$, and CD = $5 - x$.

2. Use trigonometry to form equations

Using the right triangles ABD and CBD, we can set up equations using the tangent function: $tan(60^\circ) = \frac{h}{x}$ $tan(45^\circ) = \frac{h}{5-x}$

3. Solve for h and x

We know that $tan(60^\circ) = \sqrt{3}$ and $tan(45^\circ) = 1$. Substitute these values into the equations: $\sqrt{3} = \frac{h}{x}$ so $h = x\sqrt{3}$ $1 = \frac{h}{5-x}$ so $h = 5 - x$

Since both expressions are equal to $h$, we can write: $x\sqrt{3} = 5 - x$ $x\sqrt{3} + x = 5$ $x(\sqrt{3} + 1) = 5$ $x = \frac{5}{\sqrt{3} + 1}$

4. Rationalize the denominator

Rationalize the denominator of $x$: $x = \frac{5}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{5(\sqrt{3} - 1)}{3 - 1} = \frac{5(\sqrt{3} - 1)}{2}$

5. Solve for h

Now that we have $x$, we can find $h$ using $h = 5 - x$: $h = 5 - \frac{5(\sqrt{3} - 1)}{2} = \frac{10 - 5\sqrt{3} + 5}{2} = \frac{15 - 5\sqrt{3}}{2} = \frac{5(3 - \sqrt{3})}{2}$