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

The angle $heta$ is calculated using the tangent function: $an \theta = \frac{4}{3}$ This gives a slope (m) of $\frac{3}{4}$. The ship's equation in point-slope form is given by: $y - (-10) = \frac{3}{4}(x - 0)$ Simplifying, we find the equation of the line to be: $4y = 3x - 40$ or $3x - 4y - 40 = 0$.

To find the shortest distance, we need to calculate the perpendicular distance from the lighthouse to the line described in part (b). Using the distance formula, we obtain: $d = \frac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}}$ Here, $a = 3$, $b = -4$, and $c = -40$, while the coordinates of the lighthouse $(x_1, y_1)$ are (0, 0). Plugging in these values, we find: $d = \frac{|3(0) - 4(0) - 40|}{\sqrt{3^2 + (-4)^2}} = \frac{40}{5} = 8 ext{ km}.$

The time when the ship is closest to the lighthouse can be determined using trigonometric relations. We know: $\tan \theta = \frac{8}{x}$ Where we already found $x = 6$ km. Substituting the values yields: $\frac{4}{3} = \frac{8}{x}$ Solving this gives us $x = 6$ km, which corresponds to 24 minutes after noon, making the time 12:24 pm.

Visibility is limited to 9 km. Using the equation of the line established earlier, we calculate: $y^2 + 8^2 = 9^2$
This provides us with:
$y^2 = 17, \text{ thus } y = \sqrt{17}.$
The distance travelled while visible becomes: $d = 2 \times \sqrt{17} \text{ km}.$
Since the ship travels at $15$ km/h, the time is calculated as follows:
$t = \frac{d}{v} = \frac{2 \sqrt{17}}{15} \text{ hours}$
This translates to approximately $\frac{2}{\sqrt{17}} \approx 32.98$ minutes, approximately 33 minutes.