Question

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 440m from the beacon, the sight line to the beacon makes an angle of 17° with the shoreline. what length of shoreline is effectively illuminated by the beacon (round to 1 decimal)? β = 0° answer: 0 x water 250 b 440 shoreline 17° ? shore

Explanation:

Step1: Identify the triangle

We have a triangle with sides \( OB = 440\) m, \( AB = 250\) m (where \( A \) is a point on the shoreline illuminated by the beacon), and angle at \( O \) is \( 17^\circ \). We can use the Law of Cosines to find the length of the segment on the shoreline. Let the two intersection points of the circle (radius 250 m) with the shoreline be \( P \) and \( Q \). We need to find the length \( PQ \). First, find the distance from \( O \) to the foot of the perpendicular from \( B \) to the shoreline, say \( M \). The distance \( OM = 440\cos(17^\circ) \), and the distance from \( B \) to the shoreline \( BM = 440\sin(17^\circ) \). Now, for a point \( A \) on the shoreline, the distance from \( M \) to \( A \) is \( \sqrt{250^2 - (440\sin(17^\circ))^2} \) (by Pythagoras, since \( BM \) is perpendicular to the shoreline, so triangle \( BMA \) is right-angled). Then the length \( PQ = 2\sqrt{250^2 - (440\sin(17^\circ))^2} \) (since the perpendicular from \( B \) to the shoreline bisects \( PQ \) if the shoreline is straight, but actually, we can use the Law of Cosines on triangle \( OAB \) where \( OA \) is the distance from \( O \) to \( A \), and then find the difference or sum. Wait, better to use the Law of Cosines in triangle \( OAB \): \( AB^2=OA^2 + OB^2-2\cdot OA\cdot OB\cdot \cos(17^\circ) \). Let \( OA = x \), then \( 250^2=x^2 + 440^2-2\cdot x\cdot 440\cdot \cos(17^\circ) \). This is a quadratic equation: \( x^2 - (2\cdot 440\cdot \cos(17^\circ))x + (440^2 - 250^2)=0 \).

First, calculate \( 2\cdot 440\cdot \cos(17^\circ) \approx 2\cdot 440\cdot 0.9563 \approx 841.544 \)

\( 440^2 - 250^2=(440 - 250)(440 + 250)=190\cdot 690 = 131100 \)

So the quadratic is \( x^2 - 841.544x + 131100 = 0 \)

The discriminant \( D = 841.544^2 - 4\cdot 1\cdot 131100 \approx 708196 - 524400 = 183796 \)

\( \sqrt{D} \approx 428.7 \)

Then the two roots are \( x = \frac{841.544 \pm 428.7}{2} \)

First root: \( \frac{841.544 + 428.7}{2} \approx \frac{1270.244}{2} \approx 635.122 \)

Second root: \( \frac{841.544 - 428.7}{2} \approx \frac{412.844}{2} \approx 206.422 \)

Then the length of the illuminated shoreline is the difference between these two roots: \( 635.122 - 206.422 = 428.7 \)? Wait, no, wait. Wait, the two points \( P \) and \( Q \) are on the shoreline, so the distance between them is \( |x_1 - x_2| \), where \( x_1 \) and \( x_2 \) are the distances from \( O \) to \( P \) and \( O \) to \( Q \). Wait, actually, when we solve the quadratic, the two solutions are the distances from \( O \) to the two intersection points. So the length between them is \( |x_1 - x_2| \). For a quadratic equation \( ax^2+bx + c = 0 \), the difference between roots is \( \frac{\sqrt{D}}{a} \). Here \( a = 1 \), so difference is \( \sqrt{D} \approx 428.7 \)? Wait, no, the quadratic is \( x^2 - 841.544x + 131100 = 0 \), so \( a = 1 \), \( b = -841.544 \), \( c = 131100 \). The difference between roots \( x_1 - x_2 = \frac{\sqrt{D}}{a} \) (since \( x_1=\frac{-b + \sqrt{D}}{2a} \), \( x_2=\frac{-b - \sqrt{D}}{2a} \), so \( x_1 - x_2=\frac{2\sqrt{D}}{2a}=\frac{\sqrt{D}}{a} \)). So \( \sqrt{D} \approx 428.7 \), so the length is approximately 428.7? Wait, but let's check the calculation of \( D \):

\( 440^2 = 193600 \), \( 250^2 = 62500 \), so \( 440^2 - 250^2 = 193600 - 62500 = 131100 \), correct.

\( 2\cdot 440\cdot \cos(17^\circ) \): \( \cos(17^\circ) \approx 0.956304756 \), so \( 2\cdot 440 = 880 \), \( 880\cdot 0.956304756 \approx 841.548185 \), correct.

Then \( D = (841.548185)^2 - 4\cdot 1\cdot 131100 \)

\( 841.548185^2 \approx 841.548^2 = (800 + 41.548)^2 = 800^2 + 2\cdot 800\cdot 41.548 + 41.548^2 = 640000 + 66476.8 + 1726.2 = 640000 + 66476.8 = 706476.8 + 1726.2 = 708203 \) (approx, more accurately, 841.548185*841.548185: let's calculate 841^2 = 707281, 0.548185^2 ≈ 0.3005, and cross term 2*841*0.548185 ≈ 2*841*0.548 ≈ 841*1.096 ≈ 921.736, so total ≈ 707281 + 921.736 + 0.3005 ≈ 708203.0365). Then \( 4*131100 = 524400 \), so \( D ≈ 708203 - 524400 = 183803 \), so \( \sqrt{D} ≈ \sqrt{183803} ≈ 428.7 \) (since 428^2 = 183184, 429^2 = 184041, so 428.7^2 = (428 + 0.7)^2 = 428^2 + 2*428*0.7 + 0.7^2 = 183184 + 600.4 + 0.49 = 183784.89, close to 183803, so \( \sqrt{183803} ≈ 428.72 \))

So the length of the illuminated shoreline is the distance between the two intersection points, which is the difference between the two roots of the quadratic, which is \( \frac{\sqrt{D}}{a} = \sqrt{D} ≈ 428.7 \) (wait, no, the quadratic is \( x^2 - bx + c = 0 \), so roots are \( \frac{b \pm \sqrt{D}}{2} \), so the difference is \( \frac{2\sqrt{D}}{2} = \sqrt{D} \), yes, because \( x_1 - x_2 = \frac{b + \sqrt{D}}{2} - \frac{b - \sqrt{D}}{2} = \frac{2\sqrt{D}}{2} = \sqrt{D} \)). Wait, but let's check with another approach. The distance from \( O \) to the foot of the perpendicular from \( B \) to the shoreline is \( OM = 440\cos(17^\circ) ≈ 440*0.9563 ≈ 420.772 \) m. The distance from \( M \) to \( A \) (a point on the shoreline illuminated by the beacon) is \( \sqrt{250^2 - (440\sin(17^\circ))^2} \). Calculate \( 440\sin(17^\circ) ≈ 440*0.2924 ≈ 128.656 \) m. Then \( 250^2 - (128.656)^2 = 62500 - 16552.3 = 45947.7 \), so \( \sqrt{45947.7} ≈ 214.35 \) m. Then the length \( PQ = 2*214.35 ≈ 428.7 \) m (since the perpendicular from \( B \) to the shoreline bisects \( PQ \) if the shoreline is straight, which it is, so the two segments from \( M \) to \( P \) and \( M \) to \( Q \) are equal, so total length is twice that). Yes, that matches. So the length is approximately 428.7 m.

Step2: Calculate the values

First, calculate \( \sin(17^\circ) \approx 0.2923717047 \), \( \cos(17^\circ) \approx 0.956304756 \)

Distance from \( B \) to shoreline (perpendicular): \( BM = 440 \times \sin(17^\circ) \approx 440 \times 0.29237 \approx 128.6428 \) m

Then, the length from the foot of the perpendicular \( M \) to a point \( A \) on the shoreline (illuminated) is \( \sqrt{250^2 - BM^2} = \sqrt{62500 - (128.6428)^2} \)

Calculate \( (128.6428)^2 \approx 128.6428 \times 128.6428 \approx 16550.9 \)

So \( 62500 - 16550.9 = 45949.1 \)

\( \sqrt{45949.1} \approx 214.36 \) m

Since the shoreline is straight and \( BM \) is perpendicular, the two intersection points are symmetric with respect to \( M \), so the total length \( PQ = 2 \times 214.36 \approx 428.7 \) m (rounded to 1 decimal)

Answer:

428.7