A ship is 10 km due South of a lighthouse at noon - Leaving Cert Mathematics - Question 8 - 2010

Given that $\tan(\theta) = \frac{4}{3}$, we have:

Slope $m = \frac{3}{4}$. Using the point-slope form starting from (0, -10):

[ y + 10 = \frac{3}{4}(x - 0) ] [ y = \frac{3}{4}x - 10 ]

This equation represents the path of the ship.

To find the shortest distance $d$, we can use the perpendicular distance formula:

[ d = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}} ] For the lighthouse at (0, 0) and ship's line $y = \frac{3}{4}x - 10$: Let $a = 3$, $b = -4$, and $c = -40$: [ d = \frac{|3(0) - 4(0) - 40|}{\sqrt{3^2 + (-4)^2}} ] [ d = \frac{40}{5} = 8 \text{ km} ]

The minimum distance occurs when: [ \tan(\theta) = \frac{8}{x} ] Setting this equal to $\frac{8}{x}$ and solving: [ 8 = \frac{8}{3}(x) ] Solving gives $x = 6$ km, when $y$ is: Time taken = (distance/speed) = $\frac{6}{15}$ hours which equals 24 minutes. Thus, the ship is closest to the lighthouse at 12:24 pm.

Using the right triangle visibility, we have: [ y^2 + 8^2 = 9^2 ] [ y^2 = 81 - 64 ] [ y^2 = 17 \Rightarrow y = \sqrt{17} \text{ km} ] Distance travelled from visibility point to nearest approach: [ Total time = \frac{2 \sqrt{17}}{15} \text{ hours} ] This equals [ \approx 32.98 \text{ minutes} \approx 33 minutes. ]