Applied Math- Cosine law Applied math- Cosine law

Question

Alex, Curtis and Michael go boating together on a calm day. They leave the marina dock and travel at a speed of 40 km/h due east for 3 hours. Then, they travel 22° west of south for 2 hours at a speed of 30 km/h. At that point, their boat breaks down and they call the mechanic at the marina dock to come and help them.

To get to them in the shortest time, how far must the mechanic travel and in what direction?

1. 1 mark for an appropriate diagram
2. 1 mark for finding the enclosed angle
3. 1 mark for setting cosine law up properly
4. 1 mark for finding the distances the boats travel
5. 1 mark for finding the measure of angle A
6. 1 mark for giving a correct direction

Stickman Hook

Answer 1

Step 1 Diagram (described)

1. From the marina, go 120 km due east (40 km/h × 3 h).
2. Then go 60 km at 22° west of south (30 km/h × 2 h).
3. Draw the resultant from the marina to the breakdown point.

Step 2 Enclosed angle between the two legs

1. First leg is due east.
2. Second leg is 22° west of south = direction 270°− 22° = 248°.
3. Angle between them: 112° ( since 90° + 22°).

Step 3 Set up cosine law correctly

Let a = 120 km (east leg), b = 60 km (second leg), θ = 120° the included angle

Resultant distance ccc satisfies

c² = a²+ b²+ 2 ab cos ⁡θ.

(Use “+” because it’s the magnitude of a + b with angle θ\thetaθ between them.)

Step 4 Compute distances

c² = 120² + 60² +2 (120)(60) cos⁡(112°).

Since cos⁡ (112°) = − sin ⁡(22°) ≈ − 0.3746,

c² =14400 + 3600 +14400 (−0.3746) ≈12606, c ≈ 112.3 km.

Step 5 Find ∠ A angle at the marina (direction angle)

Use the sine law:

sin⁡ A ⁄ b = sin ⁡(112°) c ⇒ sin⁡ A = 60 sin⁡ (112°) ⁄ 112.3 ⇒ A ≈ 29.7°

Step 6 Direction

From the marina, the mechanic should head 29.7° south of east ( i.e. E 29.7° S.), which is the same as a bearing of 119.7°.

Final answer

1. Shortest distance: 112.3 km (approximately)
2. Direction: E 29.7° S (bearing 119.7°)

Answer 2

Answer 3

To get to them in the shortest time, how far must the mechanic travel and in what direction?

1. 1 mark for an appropriate diagram
2. 1 mark for finding the enclosed angle
3. 1 mark for setting cosine law up properly
4. 1 mark for finding the distances the boats travel
5. 1 mark for finding the measure of angle A
6. 1 mark for giving a correct direction

No graphing support in this area of the website for problem solving, so _no_ points for me - LOL. Would be nice to paste a graphic.

• East 3h @ 40 km/h ==> East 120 km

• S22ºW 2h @ 30 km/h ==> S22ºW 60 km

  Note: This means that on the new course (after turn to starboard of 90º + 22º = 112º), there is 68º to our starboard side toward the Marina.

  68º will be the angle in our triangle between the 2 known sides: 120 km East, and 60km S22ºW.

• Our Triangle: Marina - 120 km East, Angle 68º - 60 km S22ºW, distance 'd' to Marina

•At Marina, angle 'θ' SE for Mechanic to get to the boat

For distance 'd': Law of Cosines

d² = 120² + 60² -2(120)(60)cos(68º) ==> d = √(12,606) ==> distance d=112.3 km from Marina

For Course Angle θ from Marina to the boat - Law of Sines

sin(θ) / 60 km = sin(68º) / d ==> sin(θ) = (60 km)sin(68º) / d ==> θ = sin⁻¹(0.4953787) ==> θ = 29.7º ==> Course Angle = E29.7ºS

Answer 4

construct a right triangle

due East 120 miles

then 22 degrees west of south for 2 hours at 30 mph = 60 miles

distance from start to end = d

d^2 = 120^2 +60^2 -2(60)(120cos(90-22) = 14400+3600 -5394.26= 18.000-5,394.26

d = sqr12,605.74 =about 112.28 miles

Answer 5

desmos.com/calculator/atkkej5mvc

Answer 6

Speed = distance , so speed x time = distance

Time

They travel east at 40km/hr for 3 hrs- so they travel east for a distance of 40×3=120km.

Then they travel 22 degrees west of south at a speed of 30km/hr for 2 hours. So they travel a distance of 30×2=60km. Since they travel 22 degrees west of south direction, the angle between east and west of south is 90-22=68 degrees.

So now we can use the Cosine Law because we know the lengths of two sides and the angle between them (sas).

So the distance the mechanic travels is

d ² = 120² + 60² -2(120)(60)cos68°

= 14400 + 3600 -14400cos68°

= 18000 - 5394.33

= 12605.67

d =square root of 12605.67

=112.27km

Answer 7

Distance = (Rate)(Time)

Distance travelled due east = (40)(3) = 120 miles

Distance travelled 22° west of south = (30)(2) = 60 miles

Angle formed by the two paths of travel = 90° - 22° = 68°

Let x = distance travelled by the mechanic

Using the Law of Cosines, x² = 120² + 60² - 2(120)(60)cos68° = 18000 - 14400cos68° = 18000 - 5394.24

x² = 12605.76 So, x ≈ 112.28 miles