Question

level 4 question: question 12 two search and rescue teams moved through a provincial park looking for a missing camper. search team a walked at a speed of 4.5 km per hour in a straight line 9.5° north of east, while search team b walked at a speed of 6.25 km per hour in a straight line 14.5° west of north. after 4 hours, the two teams walked toward each other, forming a triangle - shaped search area. calculate the distance between teams a and b after 4 hours of walking on their original pathways. medium/spicy extension after meeting up, teams a and b searched the full area of the triangular search pattern that they established. calculate the total area searched by teams a and b (hint: what is the formula for the area of a triangle?)

Explanation:

Question 12 (Distance between Teams A and B)

Step1: Calculate distance for Team A

Team A's speed is 4.5 km/h, time is 4 hours. Distance = speed × time, so \( d_A = 4.5 \times 4 = 18 \) km.

Step2: Calculate distance for Team B

Team B's speed is 6.25 km/h, time is 4 hours. Distance = speed × time, so \( d_B = 6.25 \times 4 = 25 \) km.

Step3: Use Law of Cosines

The angle between their paths is \( 9.5^\circ + 14.5^\circ = 24^\circ \)? Wait, no: Team A is north of east, Team B is north of west? Wait, the problem says: Team A walked at a speed of 4.5 km per hour in a straight line 9.5° north of east. Team B walked at a speed of 6.25 km per hour in a straight line 14.5° west of north. Wait, let's re-express the angle between their paths.

North of east: direction is 9.5° from east towards north. West of north: direction is 14.5° from north towards west. So the angle between the two paths: from east-north (Team A) to north-west (Team B). Let's find the angle between the two vectors.

East-north: angle with east is 9.5°, so angle with north is 90° - 9.5° = 80.5°. West of north: angle with north is 14.5° towards west. So the total angle between the two paths is \( 80.5^\circ + 14.5^\circ = 95^\circ \)? Wait, no. Let's draw the coordinate system:

• Team A: direction is 9.5° north of east. So its angle with the positive x-axis (east) is 9.5°.
• Team B: direction is 14.5° west of north. So its angle with the positive y-axis (north) is 14.5° towards west, so its angle with the positive x-axis is \( 90^\circ + 14.5^\circ = 104.5^\circ \).

The angle between the two paths is the difference between their angles with the x-axis: \( 104.5^\circ - 9.5^\circ = 95^\circ \). Wait, but maybe I made a mistake. Alternatively, the angle between east-north (9.5°) and west-north (14.5°): the angle between east and west is 180°, so from Team A's direction (9.5° north of east) to Team B's direction (14.5° west of north): let's see, the angle between east and north is 90°, Team A is 9.5° into that 90° (north of east), Team B is 14.5° into the 90° from north towards west. So the total angle between their paths is \( 90^\circ - 9.5^\circ + 14.5^\circ = 95^\circ \). Yes, that makes sense. So the angle \( \theta = 95^\circ \).

Now, using Law of Cosines: \( c^2 = a^2 + b^2 - 2ab \cos \theta \)? Wait, no: Law of Cosines is \( c^2 = a^2 + b^2 - 2ab \cos (180^\circ - \theta) \) if the angle between them is \( \theta \), but wait, no. Wait, let's re-express:

Team A's distance: \( a = 18 \) km (east-north direction)
Team B's distance: \( b = 25 \) km (west-north direction)
The angle between them: when you have east-north and west-north, the angle between the two paths is \( 9.5^\circ + 14.5^\circ + 90^\circ - 9.5^\circ - 14.5^\circ \)? No, better to use the angle between the two vectors. Let's consider the standard position (x-axis east, y-axis north).

Team A's displacement vector: \( (d_A \cos 9.5^\circ, d_A \sin 9.5^\circ) = (18 \cos 9.5^\circ, 18 \sin 9.5^\circ) \)
Team B's displacement vector: \( (-d_B \sin 14.5^\circ, d_B \cos 14.5^\circ) \) (because west of north: x-component is negative (west), y-component is positive (north))

Now, the distance between them is the magnitude of the difference between the two vectors:

\( \Delta x = 18 \cos 9.5^\circ - (-25 \sin 14.5^\circ) = 18 \cos 9.5^\circ + 25 \sin 14.5^\circ \)
\( \Delta y = 18 \sin 9.5^\circ - 25 \cos 14.5^\circ \)

Then distance \( D = \sqrt{(\Delta x)^2 + (\Delta y)^2} \)

Alternatively, use Law of Cosines with the angle between the two paths. Let's find the angle between the two paths. Team A is 9.5° north of east: so the angle between Team…

Step1: Recall Area of Triangle Formula

The formula for the area of a triangle when two sides and the included angle are known is \( \text{Area} = \frac{1}{2}ab \sin C \), where \( a \) and \( b \) are the lengths of two sides, and \( C \) is the included angle.

Step2: Identify Values

From Question 12, \( a = 18 \) km (distance of Team A), \( b = 25 \) km (distance of Team B), and the included angle \( C = 95^\circ \) (angle between their paths).

Step3: Calculate the Area

Substitute into the formula: \( \text{Area} = \frac{1}{2} \times 18 \times 25 \times \sin(95^\circ) \)
Calculate \( \sin(95^\circ) \approx \sin(90^\circ + 5^\circ) = \cos(5^\circ) \approx 0.9962 \)
So \( \text{Area} = \frac{1}{2} \times 18 \times 25 \times 0.9962 \)
First, \( 18 \times 25 = 450 \)
Then, \( \frac{1}{2} \times 450 = 225 \)
Then, \( 225 \times 0.9962 \approx 224.145 \) km²
So the total area searched is approximately 224.15 km² (or 224 km²).

Answer:

s:

• Distance between Teams A and B: Approximately 32.0 km (or 32 km).
• Total area searched: Approximately 224.15 km² (or 224 km²).