Choosing the correct trigonometric ratio - Variation Theory - Free Printable

Let's solve each of these triangle problems step by step. All the triangles are right-angled or involve a 40° angle, and we're asked to find the unknown side $ x $. We'll use trigonometric ratios (sine, cosine, tangent) depending on which sides and angles are given.

We’ll go through each problem:

---

1.

```
12cm
/|
/ |
/ | x
/___|
40°
```

This is a right triangle with:
- Hypotenuse = 12 cm
- Angle = 40°
- Opposite side = $ x $

Use sine:
$$
\sin(40^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{12}
$$
$$
x = 12 \cdot \sin(40^\circ)
$$
$$
x \approx 12 \cdot 0.6428 \approx 7.7136 \text{ cm}
$$

✔ Answer: $ x \approx 7.71 $ cm

---

2.

```
x
/|
/ |
/ | 12cm
/___|
40°
```

Right triangle:
- Adjacent = 12 cm
- Angle = 40°
- Hypotenuse = $ x $

Use cosine:
$$
\cos(40^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{x}
$$
$$
x = \frac{12}{\cos(40^\circ)} \approx \frac{12}{0.7660} \approx 15.66 \text{ cm}
$$

✔ Answer: $ x \approx 15.66 $ cm

---

3.

```
12cm
/|
/ |
/ | x
/___|
40°
```

Right triangle:
- Adjacent = 12 cm
- Angle = 40°
- Opposite = $ x $

Use tangent:
$$
\tan(40^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{12}
$$
$$
x = 12 \cdot \tan(40^\circ) \approx 12 \cdot 0.8391 \approx 10.069 \text{ cm}
$$

✔ Answer: $ x \approx 10.07 $ cm

---

4.

```
x
/|
/ |
/ | 12cm
/___|
40°
```

Right triangle:
- Opposite = 12 cm
- Angle = 40°
- Hypotenuse = $ x $

Use sine:
$$
\sin(40^\circ) = \frac{12}{x} \Rightarrow x = \frac{12}{\sin(40^\circ)} \approx \frac{12}{0.6428} \approx 18.66 \text{ cm}
$$

✔ Answer: $ x \approx 18.66 $ cm

---

5.

```
12cm
/|
/ |
/ | x
/___|
40°
```

Same as #3:
- Adjacent = 12 cm
- Opposite = $ x $
- Angle = 40°

Use tangent:
$$
x = 12 \cdot \tan(40^\circ) \approx 12 \cdot 0.8391 \approx 10.07 \text{ cm}
$$

✔ Answer: $ x \approx 10.07 $ cm

---

6.

```
12cm
/|
/ |
/ | x
/___|
40°
```

Same as #1:
- Hypotenuse = 12 cm
- Opposite = $ x $
- Angle = 40°

Use sine:
$$
x = 12 \cdot \sin(40^\circ) \approx 12 \cdot 0.6428 \approx 7.71 \text{ cm}
$$

✔ Answer: $ x \approx 7.71 $ cm

---

7.

```
x
/|
/ |
/ | 12cm
/___|
40°
```

Right triangle:
- Adjacent = 12 cm
- Angle = 40°
- Hypotenuse = $ x $

Use cosine:
$$
\cos(40^\circ) = \frac{12}{x} \Rightarrow x = \frac{12}{\cos(40^\circ)} \approx \frac{12}{0.7660} \approx 15.66 \text{ cm}
$$

✔ Answer: $ x \approx 15.66 $ cm

---

8.

```
12cm
/|
/ |
/ | x
/___|
40°
```

Same as #1 and #6:
- Hypotenuse = 12 cm
- Opposite = $ x $
- Angle = 40°

$$
x = 12 \cdot \sin(40^\circ) \approx 7.71 \text{ cm}
$$

✔ Answer: $ x \approx 7.71 $ cm

---

9.

```
14cm
/|
/ |
/ | x
/___|
12cm
```

Wait — this one has two sides labeled: 14 cm and 12 cm, and asks for $ x $. But no angle is marked. However, looking at the diagram, it seems like a right triangle with legs 14 cm and 12 cm, and $ x $ is the hypotenuse?

But wait — in the diagram, the side labeled 14 cm is horizontal, 12 cm is vertical, and $ x $ is the hypotenuse? Then it's a right triangle with legs 14 and 12.

So use Pythagoras:
$$
x = \sqrt{14^2 + 12^2} = \sqrt{196 + 144} = \sqrt{340} \approx 18.44 \text{ cm}
$$

But wait — is it really a right triangle? The angle isn't labeled as 90°. However, from the shape and labeling, it appears to be a right triangle with legs 14 and 12 cm.

Alternatively, if it's not a right triangle, we can’t solve without more info. But since all others are right triangles and this one is drawn similarly, assume it's a right triangle.

So:
$$
x = \sqrt{14^2 + 12^2} = \sqrt{196 + 144} = \sqrt{340} \approx 18.44 \text{ cm}
$$

✔ Answer: $ x \approx 18.44 $ cm

---

✔ Final Answers:

| Problem | $ x $ (cm) |
|--------|-------------------|
| 1 | ≈ 7.71 |
| 2 | ≈ 15.66 |
| 3 | ≈ 10.07 |
| 4 | ≈ 18.66 |
| 5 | ≈ 10.07 |
| 6 | ≈ 7.71 |
| 7 | ≈ 15.66 |
| 8 | ≈ 7.71 |
| 9 | ≈ 18.44 |

---

🔍 Summary of Trig Rules Used:

- Sine: $ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $
- Cosine: $ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} $
- Tangent: $ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} $
- Pythagoras: $ c = \sqrt{a^2 + b^2} $ (for right triangles)

Let me know if you'd like exact expressions (e.g., $ x = 12 \sin(40^\circ) $) instead of decimals!