**TASK 1**

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Assumption: The men have no height. They are looking at the same point of the building.Let the distance of the building from the man with the 30° angle be y, and the distance from the man with the 60° angle by x.

**CASE 1:**

Both men facing the same direction:

tan30°=120/y

y=120/tan30°

y=208m (3sf)

tan60°=120/x

x=120/tan60°

x=69.3m (3sf)

208-69.3=139m (3sf)

**CASE 2:**

Both men facing different directions:

y=120/tan30°

y=208m (3sf)

x=120/tan60°

x=69.3m (3sf)

208+69.3=277m (3sf)

**TASK 2**

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**SINE RULE**

Draw a line h that is perpendicular to line AB

It will form 2 right angled triangles

BC = a

AC = b

AC = b

So…

Sin A = h/b

Sin B = h/a

Sin B = h/a

h = bSin A

h = aSin B

h = aSin B

bSin A = aSin B

Divide both sides by ab

Sin A/a = Sin B/b

If we do it on another side we can find out that

Sin C/c = Sin A/a = Sin B/b

Remember ASTC, the sine value is still positive in the 2nd quadrant.

sin A =sin180-A

For example:

The sin of 20º = sin 180º-20º

The value are the same even though the angles are different

So that means that there can be 2 solutions to a problem which results in this being the ambiguous case.

sin A =sin180-A

For example:

The sin of 20º = sin 180º-20º

The value are the same even though the angles are different

So that means that there can be 2 solutions to a problem which results in this being the ambiguous case.

**COSINE RULE**

First we divide the ∆ABC into two using its height. Note that the side is named after the angle opposite it. (A is the angle, and the side directly opposite it is a)

Using angle A, cosA = x/b and therefore

0. bcosA = x

Treating the base of the smaller triangle as 'x', the longer base becomes c-x

Now treating both sides a and b as hypotenuses of their respective right angled ∆s, we use Pythagoras' Theorem to derive the following:

1. b^2 = h^2 + x^2

2. a^2 = h^2 + (c - x)^2

And by manipulating (2), we get:

3. x^2 + h^2 = a^2 - c^2 + 2cx

Substitute (1) into (3),

4. b^2 = a^2 - c^2 + 2cx

Substitute (0) into (4)

b^2 = a^2 - c^2 +2bcosA

5. b^2 + c^2 - 2bcosA = a^2

(5) can be rearranged to get:

cosA = (b^2 + c^2 - a^2)/2bc

Subsequently,

cosB = (a^2 + c^2 - b^2)/2ac

**TASK 3**

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**HEURISTIC PROBLEM SOLVING**

Assumptions: The girl is moving on a flat plane. The directions are in respect to true north.

ACTIVITY 1:

A: The maximum displacement of the particles from the rest position

B:The shortest distance between any two points in a wave that are in phase

C: The number of complete waves produced per second

D: Phase is sinusoidal functions

E: The amplitude, wavelength, speed and phase constant of the wave

ACTIVITY 2:

Increase in amplitude

Similar

y=3sinx

Increase in frequency

Similar

y=sin2x

Shifted to the right

Similar

y=sin(x-1)

Shifted up

Similar

y=sinx + 3

A visual representation would be much appreciation. Please include a diagram.

ReplyDeleteFor the second solution, they did not assume that the building has no thickness.

ReplyDelete.

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