Heuristic Problem Solving - Trigonometry
TASK 1
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Two men facing a tall building notice the angle of elevation to the top of the building to be 30˚ and 60˚ respectively. If the height of the building is known to be h =120 m, find the distance (in meters) between the two men.
Assumptions:
- 2 men are standing on one side of the building, lying on the same, flat plane
- both men are of the same height and looking at the same point of the building
- both men are in the same medium (air)
- the building is in vertical position and perpendicular to the ground
Solving the problem:
Let 'a' be the distance between the first man and the building (30˚ elevation)
Let 'b' be the distance between the second man and the building (60˚ elevation)
Let 'x' be the distance between the two men
tan60˚ = 120 ÷ a
a = 120 ÷ tan60˚
tan30˚ = 120 ÷ b
b = 120 ÷ tan30˚
x = (120 ÷ tan30˚) - (120 ÷ tan60˚)
= 138.6
= 139 m (3sf)
∴ the answer is 139 m
(17/7/13)
Trigonometry - Sine and Cosine
TASK 2
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Sine Rule
Basically:
Cosine = Adjacent / hypotenuse
Cosine is a ratio
Pythagoras theorem on the left triangle (ACD) is BLUE
Pythagoras theorem on the right triangle (BCD) is RED.
Assuming that a, b and c are interchangeable due to it being points of a triangle.
The cosine of 90º is 0.
Making the last formula a^2 + c^2 = b^2 as 0 times anything = 0.
Trigonometry - Bearing
TASK 3
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A triangular paddock has two complete fences. From location D, one fence line is on a bearing of N23˚W for 400 metres. The other fence line is S55˚W for 700 metres.
Find the length of fencing (to the nearest metre) required to complete the enclosure of the triangular paddock.
Assumptions:
- All fences are on the same plane
Solving the problem:
draw a diagram to be able to understand more clearly on what the question is about.
From the diagram, you are able to find an include angle in the triangle.
As the point is touching on a straight line, thus the angles on the straight line, adds up to 180˚.
∴ 180˚ - 23˚ - 55˚ = 102˚
and then, we can take the triangle.
from this, we can work out the problem by using cosine rule.
let the required length to be 'c'
c² = 400² + 700² - 2(400)(700)cos102˚
c = 875 m (nearest metre)
∴ the answer is 875 m
Question 2:
Soldiers on a reconnaissance set off on a return journey from their base camp. The journey consists of three legs. The first leg is on a bearing of 150˚T for 3 km; the second is on a bearing of 220˚T for 5 km.
Find the direction and distance of the third leg by which the group returns to its base camp.
Solving the problem:
finding the direction:
sin30˚ ÷ 5 km = sin(3rd leg) ÷ 3 km
0.3 = sin(3rd leg)
3rd leg = 17.5˚ (1dp)
finding the distance:
sin132.5 ÷ sin30˚ ÷ 51 cm = 7.37 cm
Trigonometry - Graph Transformation
TASK 4
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Activity 1
A. Amplitude - the amplitude of a wave is its maximum disturbance from its undisturbed position
B. Wavelength - the length of a complete wave is from a crest to a crest, a trough to a trough, meaning its one point to the next consecutive point
C. Frequency - is the number of waves occurring in 1 second
D. Phase - the relationship in time between the successive cycles or states of an oscillating or repeating system and either a fixed reference point or the cycles or states of another system with which it may or may not be in synchrony.
E. Characteristics of the Wave - they are the amplitude, the wavelength, frequency and phase.
Activity 2
Transformation 1: y = 3sin(x)
Transformation 2: y = sin(2x)
Transformation 3: y = sin(x - c)
(increases the x value by c thus, if c is positive, the graph moves to the left. If c is negative, the graph moves to the right)
Transformation 4: y = sin(x) + 3
done by: G, J, Y
Trigonometry - Graph Transformation
Activity 1
Amplitude : The distance between the crest and trough divided by two.
Wavelength: The length of one complete wave or oscillation
Frequency: The number of waves occurring per cycle.
Phase: Two points that are in successive cycles away from each other.
Characteristics of the wave: The values of the amplitude, wavelength, frequency or period.
Activity 2
Activity 2
A diagram illustrating the various values of the sine graph
The graph y=sin(x) and y=3sin(x). Vertically stretched
The graph y=sin(x) and y=sin(x) + 4. Elevated by 4 units
The graph y=sin(x) and y=sin(x-1). Shifted to the right of the x-axis
The graph y=sin(x) and y=sin(2x). Compressed horizontally by factor of two.
Done by: SDS
If the 2 men are looking at eye level, then does that mean that the height below eye level is negligible?
ReplyDeleteEven if the two men are looking at the same building, was it during the same time? e.g. the building could have renovated during the time interval between the men?
Other than that, the explanation was clear and colorful. Keep Up the GREAT work ^w^!
~Esther, Group 4 Noch Noch
Would the height of the men be neglible, or is the height, 120m take into account the height of the man.
ReplyDeleteIs the building the same height from all points?
Their explanation is good and their answer is precise.
Height cannot be negligible, unless they're on the same eye level, like the taller guy is on a step and looking at the building, then since the question only said distance, i assume its horizontal distance and not the shortest distance since it would be diagonal. We assume that the building is not renovated. If both men were on the same flat surface as the building, then the height of the building that the men saw should not be 120m, since their own height has to be taken into account, this is why workers use a special tool to measure the height of the building.
ReplyDeleteGood analysis from the team but do take into cognizance of the comments made.
ReplyDeleteThe key issue is about practicality and appropriateness of solution based on the assumptions made.