HCPSS Worthwhile Math Task
The student will demonstrate the ability to use a problem-solving approach in exploring the properties of vectors and applications of parametric equations.
Find the norm (or magnitude) and direction of a geometric vector.
Use vectors to model and solve real-world problems, including velocity, force, and air navigation.
MP1: Make sense of problems and persevere in solving them.
MP3: Construct viable arguments and critique the reasoning of others.
MP4: Model with mathematics.
MP5: Use appropriate tools strategically.
MP6: Attend to precision.
MP7: Look for and make use of structure.
Trigonometry Honors, Unit 1
A tour boat takes sightseeing passengers along a river for a two hour trip. At the end of the trip, the passengers board a bus to continue the tour. The tour boat captain, Willie, is directed by the owner of the company to travel directly downstream at a speed of 28 miles per hour in order to maintain optimal viewing of the shoreline. The boat is traveling with a current of 5 miles per hour 10 ᵒ east of south. In what direction and speed does Willie need to pilot the boat in order to maintain the owner’s specifications?
1.
Students will need calculators for this task. (Look for evidence of MP5.)
2.
Assign students to pairs to allow for discussion of the scenario. (Look for evidence of
MP1 and MP3.)
3.
As the students work together, check diagrams to be sure given information is labeled properly. (Look for evidence of MP4 and MP6.)
4.
Students should use formulas for calculating magnitude and direction angles. (Look for evidence of MP7.)
Howard County Public Schools Office of Secondary Mathematics Curricular Projects has licensed this product under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0
Unported License .

HCPSS Worthwhile Math Task
5.
This task can be solved using Law of Cosines from Unit 0. Encourage the students to solve using the vector method, then go back and check their answers using the Law of
Cosines.
1.
If the current is flowing upstream (5 miles per hour 10 ᵒ east of north) instead of downstream, in what direction and speed does Willie need to pilot the boat in order to maintain the owner’s specifications?
2.
Why is the speed required for Follow-Up Question 1 faster than the speed required for the original situation?
28 mph
10 ᵒ
5 mph 𝑥 2 𝑥 2
= 28 2 + 5 2
= 533.254
𝑥 = 23.092
− 2(28)(5) cos(10)
(23.092
2 + 5 2 − 28 2 𝑐𝑜𝑠⍬ =
(2)(23.092)(5)
⍬ = 𝑐𝑜𝑠 −1 (−0.978)
)
⍬ = 167.887°
180 − 10 − 167.887 = 2.113°
= −0.978
Willie needs to pilot the boat 23.092 miles per hour. The direction angle for piloting should be
267.887
ᵒ or 2.113
ᵒ west of south.
For Math Analysis, or PCGT, this problem can also be solved using algebraic vectors:
Actual Boat Path Vector <0, -28>
Howard County Public Schools Office of Secondary Mathematics Curricular Projects has licensed this product under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0
Unported License .

HCPSS Worthwhile Math Task
Current Path Vector <5cos 280 ᵒ , 5sin 280 ᵒ >
Boat Piloting Vector <-5cos 280 ᵒ , -28 – 5sin 280 ᵒ >
Boat Piloting Vector Magnitude = (

5 cos 280

)
2 
(

28

5 sin 280

)
2 
23 .
092
Boat Piloting Direction tan
 

28

5

5 cos sin 280
280



87 .
845

Since the direction angle is in quadrant 3, the direction angle equals 87.845 + 180 = 267.845
ᵒ or 2.155
ᵒ west of south.
Follow Up Questions:
1.
10 ᵒ
5 mph
Note: Figure not drawn to scale.
28 mph 𝑥 2 𝑥 2
= 28 2 + 5 2 − 2(28)(5) cos(170)
= 1084.746
𝑥 = 32.935
(32.935
2 + 5 2 − 28 2 𝑐𝑜𝑠⍬ =
(2)(32.935)(5)
⍬ = 𝑐𝑜𝑠 −1 (0.989)
⍬ = 8.506
°
)
10 − 8.506 = 1.494°
= 0.989
Willie needs to pilot the boat 32.935 miles per hour. The direction angle for piloting should be
268.506
ᵒ or 1.494
ᵒ west of south.
For Math Analysis, or PCGT, this problem can also be solved using algebraic vectors:
Actual Boat Path Vector <0, -28>
Howard County Public Schools Office of Secondary Mathematics Curricular Projects has licensed this product under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0
Unported License .

HCPSS Worthwhile Math Task
2.
Current Path Vector <5cos 80 ᵒ , 5sin 80 ᵒ >
Boat Piloting Vector <-5cos 80 ᵒ , -28 – 5sin 80 ᵒ >
Boat Piloting Vector Magnitude = (

5 cos 80

)
2 
(

28

5 sin 80

)
2 
32 .
935
Boat Piloting Direction tan
 

28

5

5 sin 80 cos 80



88 .
489

Since the direction angle is in quadrant 3, the direction angle equals 88.489 + 180 = 268.489
ᵒ or 1.511
ᵒ west of south.
Since the current in Follow-Up question 1 is going against the path of the boat, the captain needs to push the boat with a greater force to overcome the current. In the original question, the current travels downstream with the boat which helps push the boat in the right direction, therefore the boat does not need to push through the water with as much force as Follow-Up question 1.
Howard County Public Schools Office of Secondary Mathematics Curricular Projects has licensed this product under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0
Unported License .