Worksheet featuring trigonometry application problems requiring diagramming and calculation, with a reverse illustration table and numerical data.

Let's solve each of the trigonometry application problems step by step, following the instructions: read carefully, draw a diagram (mentally or on paper), identify right triangles, apply trigonometry, and round to the nearest whole number.

---

C. A rectangle is 100 metres long and 50 metres wide. Find the acute angle between the diagonals.

#### Step 1: Understand the Geometry
- Rectangle with length = 100 m, width = 50 m.
- Diagonals of a rectangle are equal and bisect each other.
- The diagonals intersect at the center, forming two pairs of vertical angles, and we want the acute angle between them.

Let’s consider triangle formed by half-diagonals and one side.

#### Step 2: Use Coordinates or Triangle
Let’s place the rectangle with corners at:
- $ A(0,0), B(100,0), C(100,50), D(0,50) $

Diagonals: AC and BD.

The diagonals intersect at the midpoint $ M(50, 25) $.

Now consider triangle formed by vectors from the center to two adjacent corners:
- From $ M $ to $ A(0,0) $: vector $ \vec{MA} = (-50, -25) $
- From $ M $ to $ B(100,0) $: vector $ \vec{MB} = (50, -25) $

But easier: think of triangle formed by half-diagonals and half-width/length.

Alternatively, use geometry:

Each diagonal splits the rectangle into two right triangles. The diagonals cross and form four angles at the center.

We can find the angle between the diagonals using vectors or trigonometry.

Let’s use triangle formed by half-length, half-width, and half-diagonal.

At the intersection point, the diagonals split into four segments. Consider the triangle formed by:
- Half of length = 50 m
- Half of width = 25 m
- This forms a right triangle with legs 50 and 25?

Wait — actually, better way:

Let’s consider triangle formed by:
- The two diagonals intersect at center.
- Each diagonal has length $ d = \sqrt{100^2 + 50^2} = \sqrt{10000 + 2500} = \sqrt{12500} = 50\sqrt{5} \approx 111.8 $ m

But we don’t need full diagonal.

Instead, consider the triangle formed at the center: from center to corner, and to adjacent corners.

Actually, simpler method:

Let’s look at the triangle formed by:
- The two diagonals intersect at the center.
- They form an "X" shape.
- The acute angle between them can be found by considering the triangle formed by half the rectangle.

Let’s take the triangle from the center to the midpoints of sides.

Better idea: Use vector dot product or right triangle approach.

Alternate method:

In rectangle, diagonals bisect each other. So, consider triangle made by:
- Half the diagonal: from center to corner.
- But instead, let's use the fact that the diagonals create two congruent right triangles.

Actually, here’s a better way:

Let’s look at the triangle formed by:
- The two diagonals intersect at the center.
- The angle between them can be found by considering the triangle formed by:
- One half of the diagonal
- Half of the length
- Half of the width

But wait — actually, if you draw both diagonals, they intersect at the center. Then, the angle between them can be found using the slopes.

Let’s do this:

Slope of diagonal from $ (0,0) $ to $ (100,50) $: $ m_1 = \frac{50}{100} = 0.5 $

Slope of diagonal from $ (0,50) $ to $ (100,0) $: $ m_2 = \frac{0 - 50}{100 - 0} = -0.5 $

Angle $ \theta $ between two lines with slopes $ m_1 $ and $ m_2 $:

$$
\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
= \left| \frac{0.5 - (-0.5)}{1 + (0.5)(-0.5)} \right|
= \left| \frac{1}{1 - 0.25} \right| = \frac{1}{0.75} = \frac{4}{3}
$$

So,
$$
\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ
$$

This is the acute angle between the diagonals.

✔ Answer: 53°

---

E. A marksman aims at a target bullseye 60 metres away but just before he pulls the trigger he is distracted and swings the rifle and angle of a quarter of a degree. By how many centimetres will the bullet miss the bullseye?

#### Step 1: Diagram
- Marksman shoots at 60 m distance.
- Rifle is swung off by $ 0.25^\circ $ (quarter of a degree).
- We assume the bullet travels in a straight line along the new direction.
- So, it misses the bullseye by a lateral distance.

This is a small angle problem.

Use tangent:
$$
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{60}
$$
where $ x $ is the miss distance.

$ \theta = 0.25^\circ $

So:
$$
x = 60 \cdot \tan(0.25^\circ)
$$

Calculate:
$$
\tan(0.25^\circ) \approx 0.004363
$$
$$
x \approx 60 \times 0.004363 = 0.2618 \text{ metres} = 26.18 \text{ cm}
$$

Round to nearest whole number: 26 cm

✔ Answer: 26 cm

---

I. A gable roof (isosceles triangle-shaped) has a vertical height of 2.1 metres and the ceiling is 10.9 metres from one side to the other. Find the pitch (angle) of the roof.

#### Step 1: Understand
- Gable roof: isosceles triangle.
- Height = 2.1 m (from base to peak).
- Base = 10.9 m (total width).
- So, half-base = $ \frac{10.9}{2} = 5.45 $ m

We want the pitch, which is the angle between the sloped side and the horizontal (or sometimes the angle at the peak — but usually it's the angle of slope).

Assuming pitch means the angle between the roof slope and the horizontal.

So, in the right triangle:
- Opposite = height = 2.1 m
- Adjacent = half-base = 5.45 m
- Angle $ \theta $ at base

$$
\tan(\theta) = \frac{2.1}{5.45} \approx 0.3853
$$
$$
\theta = \tan^{-1}(0.3853) \approx 21.1^\circ
$$

Round to nearest whole number: 21°

✔ Answer: 21°

---

P. A searchlight shines on a flying saucer vertically above. Another searchlight shines on the saucer but its beam makes an angle of 15° to the vertical. The lights are 2 kilometers apart. Find the altitude of the flying saucer, in metres.

#### Step 1: Diagram
- Two searchlights, 2 km apart.
- One shines vertically up to saucer → so the saucer is directly above that light.
- Second light shines at 15° to vertical and hits the same saucer.

So, the saucer is directly above the first light, and the second light must aim at an angle of 15° from vertical to reach the saucer.

Let’s define:
- Let $ h $ be the altitude (height) of the saucer.
- Distance between lights = 2 km = 2000 m
- From second light, the beam goes to saucer at 15° from vertical.

So, in the triangle:
- Vertical leg = $ h $
- Horizontal leg = 2000 m
- Angle between vertical and hypotenuse = 15°

So, horizontal leg = $ h \cdot \tan(15^\circ) $

Because:
$$
\tan(15^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{2000}{h}
\Rightarrow h = \frac{2000}{\tan(15^\circ)}
$$

Now:
$$
\tan(15^\circ) \approx 0.2679
$$
$$
h = \frac{2000}{0.2679} \approx 7464.2 \text{ m}
$$

Round to nearest whole number: 7464 m

✔ Answer: 7464 m

---

R. A line on a graph joins the origin to the point (6,15). What angle does the line make with the x-axis?

#### Step 1: Use tangent
- The angle $ \theta $ with x-axis:
$$
\tan(\theta) = \frac{y}{x} = \frac{15}{6} = 2.5
$$
$$
\theta = \tan^{-1}(2.5) \approx 68.2^\circ
$$

Round to nearest whole number: 68°

✔ Answer: 68°

---

T. A regular octagon with sides of 10 metres fits exactly inside a circle. Find the radius of the circle.

#### Step 1: Regular Octagon Inscribed in Circle
- All vertices lie on the circle.
- Side length = 10 m
- We want the radius $ R $

For a regular polygon with $ n $ sides, side length $ s $, radius $ R $:

$$
s = 2R \sin\left(\frac{\pi}{n}\right)
$$

Here, $ n = 8 $, $ s = 10 $

$$
10 = 2R \sin\left(\frac{\pi}{8}\right)
$$

$$
\sin\left(\frac{\pi}{8}\right) = \sin(22.5^\circ) \approx 0.3827
$$

$$
10 = 2R \cdot 0.3827 \Rightarrow R = \frac{10}{2 \cdot 0.3827} = \frac{10}{0.7654} \approx 13.06 \text{ m}
$$

Round to nearest whole number: 13 m

✔ Answer: 13 m

---

U. A child sits on a swing which is not moving. The rope is 400 centimetres long and the child’s seat is 50 centimetres above the ground. The child swings and at the extremity of her movement the rope makes an angle of 41° with the vertical. How far above the ground is her seat now?

#### Step 1: Understand
- Rope length = 400 cm
- Initial height of seat = 50 cm above ground
- At maximum swing, rope makes 41° with vertical.
- We want new height of seat above ground.

When the swing moves, the rope remains taut, so the vertical component of the rope changes.

Let’s compute how much lower the seat is.

From the pivot point (top of rope), the rope is 400 cm long.

At extreme position, the rope makes 41° with vertical.

So, the vertical drop from pivot to seat is:
$$
\text{Vertical component} = 400 \cdot \cos(41^\circ)
$$

We need to compare this to the original vertical component when hanging straight down.

Originally, vertical component = 400 cm (full length)

Now, vertical component = $ 400 \cos(41^\circ) $

So, the drop from lowest point is:
$$
\Delta h = 400 - 400 \cos(41^\circ) = 400(1 - \cos(41^\circ))
$$

But we need new height above ground.

Original seat height = 50 cm above ground.

But the pivot point is at height: $ 50 + 400 = 450 $ cm above ground.

Now, at extreme swing, the seat is lowered by:
$$
\text{Vertical displacement} = 400 - 400 \cos(41^\circ)
$$

So, new height of seat:
$$
h = 450 - (400 - 400 \cos(41^\circ)) = 450 - 400 + 400 \cos(41^\circ) = 50 + 400 \cos(41^\circ)
$$

Wait — no:

Actually:
- Pivot is at 450 cm
- Seat is now at vertical distance $ 400 \cos(41^\circ) $ below pivot
- So seat height = $ 450 - 400 \cos(41^\circ) $

Compute:
$$
\cos(41^\circ) \approx 0.7547
$$
$$
400 \cdot 0.7547 = 301.88
$$
$$
\text{Seat height} = 450 - 301.88 = 148.12 \text{ cm}
$$

Round to nearest whole number: 148 cm

✔ Answer: 148 cm

---

Summary of Answers:

| Problem | Answer |
|--------|--------|
| C | 53° |
| E | 26 cm |
| I | 21° |
| P | 7464 m |
| R | 68° |
| T | 13 m |
| U | 148 cm |

---

Now, match these answers to the Illustration in reverse table:

Given numbers:
26, 68, 148, 13, 53, 21, 7464

These correspond to:

- E: 26
- R: 68
- U: 148
- T: 13
- C: 53
- I: 21
- P: 7464

So, the illustration in reverse likely means that these numbers go into the boxes in order of the letters.

But the table has:

```
BHC
2008
0049
[ 26 ] [ 68 ] [ 148 ] [ 13 ] [ 53 ] [ 21 ] [ 7464 ]
```

And blank row below.

Possibly, the answers are to be filled in order of the letter labels: C, E, I, P, R, T, U

But the list given is: C, E, I, P, R, T, U

Wait — the problems are labeled: C, E, I, P, R, T, U

So order is: C, E, I, P, R, T, U

Answers:
- C: 53
- E: 26
- I: 21
- P: 7464
- R: 68
- T: 13
- U: 148

So fill the blanks in order:

| C | E | I | P | R | T | U |
|---|---|---|---|---|---|---|
| 53 | 26 | 21 | 7464 | 68 | 13 | 148 |

But the table shows:

```
[ 26 ] [ 68 ] [ 148 ] [ 13 ] [ 53 ] [ 21 ] [ 7464 ]
```

Which is E, R, U, T, C, I, P

So maybe the illustration is scrambled.

But the instruction says "Illustration in reverse", perhaps meaning the answers are listed in reverse order.

Alternatively, it might be a cipher or code.

But since all problems are solved, and answers match the numbers given, we can conclude:

✔ All problems solved correctly.

Final Answers:

- C: 53°
- E: 26 cm
- I: 21°
- P: 7464 m
- R: 68°
- T: 13 m
- U: 148 cm

These match the numbers in the table: 26, 68, 148, 13, 53, 21, 7464 — just not in order.

So the correct order of answers is:

53, 26, 21, 7464, 68, 13, 148

But the table has them as: 26, 68, 148, 13, 53, 21, 7464

So perhaps the table is not in order of the problems.

But since the task was to solve the problems, we have done so.

---

✔ Final Answer Summary:

| Problem | Answer |
|--------|----------------|
| C | 53° |
| E | 26 cm |
| I | 21° |
| P | 7464 m |
| R | 68° |
| T | 13 m |
| U | 148 cm |

All correct to nearest whole number.