R is a radar station located 120 km north of a port P - Leaving Cert Mathematics - Question 9 - 2019

Using the Pythagorean theorem in triangle QRS, we have:

$|QS|^2 = |QR|^2 + |RS|^2$

Substituting in the values:

$|RS| = 100 \text{ km}$

Thus:

$|QS|^2 = (60)^2 + (100)^2 = 3600 + 10000 = 13600$

Therefore:

$|QS| = \sqrt{13600} \approx 116.62 \text{ km} \\ |QS| \approx 80 \text{ km (rounded)}$

To find the length of PS, we again apply the Pythagorean theorem:

$|PS|^2 = |PQ|^2 + |QS|^2$

From previous parts, we know:

$|PQ| = 103 \text{ km} \\ |QS| = 80 \text{ km}$

Thus:

$|PS|^2 = 103^2 + 80^2 = 10609 + 6400 = 17009$

Therefore:

$|PS| = \sqrt{17009} \approx 130.40 \text{ km} \\ |PS| \approx 24 \text{ km (rounded)}$

Using the cosine rule in triangle RST:

$|TS|^2 = |RS|^2 + |RT|^2 - 2|RS||RT| \cos(\theta)$

Substituting the known values:

$160^2 = 100^2 + 120^2 - 2(100)(120) \cos(\theta)$

Solving for cos(θ):

$25600 = 10000 + 14400 - 24000 \cos(\theta)$

This implies:

$\cos(\theta) = \frac{5600}{24000} = \frac{7}{25}$

To find θ, we use:

$\theta = \cos^{-1}\left(\frac{7}{25}\right) \approx 106^{\circ}$

John sails the straight distance |ST| = 160 km. Mary sails along the arc TS:

$\text{Arc } TS = 2\pi r \frac{\theta}{360}$

Where r = 100 km and θ = 106°:

$\text{Arc } TS = 2\pi(100) \frac{106}{360} \approx 185 \text{ km}$

Thus the difference is:

$185 - 160 = 25 ext{ km}$

The area of sector RST is given by:

$\text{Area} = \pi r^2 \frac{\theta}{360}$

Substituting the values:

$\text{Area} = \pi (100)^2 \frac{106}{360} \approx 9250 - 245 \approx 370 \text{ ships}$