Question

1. the light from a rotating offshore beacon can illuminate effectively up to a distance of 250 m. from a point on the shore that is 500 m from the beacon, the sight line to the beacon makes an angle of 20° with the shoreline. what length of shoreline is effectively illuminated by the beacon? (i.e. solve for the length of ad in the diagram below.)

Explanation:

Step1: Use the Law of Sines

In \(\triangle ABC\), by the Law of Sines \(\frac{AC}{\sin\angle ABC}=\frac{AB}{\sin\angle ACB}\). Given \(AB = 250\), \(BC\) is the distance from beacon to \(C\) (\(BC=\sqrt{500^{2}+250^{2}}\) using Pythagoras in right - triangle with legs \(500\) and \(250\), but Law of Sines is better here). Wait, no, in \(\triangle ABC\), \(\angle ACB = 20^{\circ}\), \(AB = 250\), and in \(\triangle BCD\) (right - triangle with \(BD\) part of shoreline and \(B\) to \(D\) via \(250\) m arc). Wait, another approach:
Let's consider the angles. The angle between \(BC\) and the perpendicular from \(B\) to shoreline is \(60^{\circ}\).
We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\).
In the right - triangle formed by the perpendicular from \(B\) to shoreline (\(500\) m) and the segments on the shoreline.
The total length of the illuminated shoreline \(AD\):
We use the formula for the length of a chord in a circle (but here using trigonometry).
Let's consider two right - triangles.
For the left - hand side (from the perpendicular from \(B\) to \(C\) part): \(\tan60^{\circ}=\frac{l_1}{500}\), so \(l_1 = 500\tan60^{\circ}\)
For the right - hand side (from the perpendicular from \(B\) to \(D\) part): \(\tan(60 - 40)^{\circ}=\tan20^{\circ}=\frac{l_2}{500}\) (Wait, no, correct approach:
We know that if we consider the two triangles on either side of the perpendicular from \(B\) to shoreline.
The angle between \(BC\) and the perpendicular is \(60^{\circ}\), and the angle between \(BD\) and the perpendicular is \(60^{\circ}\) (because of the symmetry of the \(250\) m arcs).
The length of \(AD\) can be found using \(\tan\alpha=\frac{\text{length from perpendicular}}{\text{500}}\)
The total length \(AD = 2\times500\tan60^{\circ}\)
Since \(\tan60^{\circ}=\sqrt{3}\)
\(AD = 1000\sqrt{3}\approx1000\times1.732 = 1732\) m

Another correct trigonometric approach (using the Law of Sines in \(\triangle ABC\) and \(\triangle ABD\)):
In \(\triangle ABC\), \(\angle BAC = 180-(60 + 20)=100^{\circ}\), \(\angle ABC = 20^{\circ}\), \(AB = 250\), \(BC\) is calculated via Law of Sines \(\frac{BC}{\sin100^{\circ}}=\frac{250}{\sin20^{\circ}}\), \(BC=\frac{250\sin100^{\circ}}{\sin20^{\circ}}\)
But the best way:
We know that the two non - overlapping right - triangles (on either side of the perpendicular from \(B\) to shoreline) have an angle of \(60^{\circ}\) (because the angle between the light ray and the perpendicular is \(60^{\circ}\) (from the \(60^{\circ}\) given in the diagram)).
The length from the perpendicular to \(C\) (or \(D\)) is \(x = 500\tan60^{\circ}\)
Since \(AD=2x\)
\(AD = 2\times500\times\sqrt{3}=1000\sqrt{3}\approx1732\) m

Answer:

\(1732\) m