Question

1. the observation deck of the skylon tower in niagara falls, ontario, is 166 m above the niagara river. a tourist in the observation deck notices two boats on the river. from the tourist’s position, • the bearing of boat a is 40° • the bearing of boat b is 258° • the angle of depression to boat b is 34°. calculate the distance between the two boats to the nearest metre. (286 m)

1. two roads intersect at 34°. two cars leave the intersection on different roads at speeds of 90 km/h and 100 km/h. after 2 h, a traffic helicopter that is above and between the two cars takes readings on them. the angle of depression to the slower car is 20° and the straight-line distance from the helicopter to that car is 100 km. assume that both cars are travelling at constant speed. sketch.

a) calculate the straight-line distance, to the nearest kilometre, from the helicopter to the faster car. (39 km)
b) determine the altitude of the helicopter to the nearest kilometre. (34 km)

1. simone is facing north at the entrance of a tunnel through a mountain. she notices that a 1515 m high mountain in the distance has a bearing of 270° and its peak appears at an angle of elevation of 35°. after she exits the tunnel, the same mountain has a bearing of 258° and its peak appears at an angle of elevation of 31°. assuming that the tunnel is perfectly level and straight, how long is it to the nearest metre? (605 m)

1. an airport radar operator locates two planes flying toward the airport. the first plane, p, is 120 km from the airport, a, at a bearing of 70° and with an altitude of 2.7 km. the other plane, q, is 180 km away on a bearing of 125° and with an altitude of 1.8 km. calculate the distance between the two planes to the nearest tenth of a kilometre. (148.4 km)

Explanation:

To solve problem 14 (distance between two planes), we use the Law of Cosines for the horizontal distance and then the Pythagorean Theorem for the 3D distance.

Step 1: Find the horizontal angle between the planes

The bearing of plane \( P \) is \( 70^\circ \), and the bearing of plane \( Q \) is \( 125^\circ \). The angle between their horizontal paths is:

$$ \theta = 125^\circ - 70^\circ = 55^\circ $$

Step 2: Calculate the horizontal distance (\( d_{\text{horizontal}} \)) between the planes

Using the Law of Cosines in the horizontal plane (triangle with sides \( 120 \, \text{km} \), \( 180 \, \text{km} \), and included angle \( 55^\circ \)):

$$ d_{\text{horizontal}}^2 = 120^2 + 180^2 - 2(120)(180)\cos(55^\circ) $$

First, compute \( \cos(55^\circ) \approx 0.5736 \):

$$ d_{\text{horizontal}}^2 = 14400 + 32400 - 2(120)(180)(0.5736) $$

$$ d_{\text{horizontal}}^2 = 46800 - 24725.76 = 22074.24 $$

$$ d_{\text{horizontal}} \approx \sqrt{22074.24} \approx 148.57 \, \text{km} $$

Step 3: Calculate the vertical distance (\( d_{\text{vertical}} \)) between the planes

The altitude of \( P \) is \( 2.7 \, \text{km} \), and the altitude of \( Q \) is \( 1.8 \, \text{km} \). The vertical distance is:

$$ d_{\text{vertical}} = 2.7 - 1.8 = 0.9 \, \text{km} $$

Step 4: Calculate the 3D distance (\( d_{\text{total}} \)) between the planes

Using the Pythagorean Theorem (horizontal and vertical distances are perpendicular):

$$ d_{\text{total}}^2 = d_{\text{horizontal}}^2 + d_{\text{vertical}}^2 $$

$$ d_{\text{total}}^2 \approx (148.57)^2 + (0.9)^2 \approx 22073.0 + 0.81 \approx 22073.81 $$

$$ d_{\text{total}} \approx \sqrt{22073.81} \approx 148.4 \, \text{km} $$

Answer:

The distance between the two planes is \(\boldsymbol{148.4 \, \text{km}}\) (to the nearest tenth of a kilometre).