Constructing Triangles - Go Teach Maths: Handcrafted Resources for ... - Free Printable

Let's solve each of the nine triangle problems step by step, using triangle construction principles and trigonometry (specifically the Law of Sines and Law of Cosines) where needed. The task is to construct each triangle and measure or calculate the missing side or angle.

We’ll go through each one:

---

1. Triangle with angles 45°, 75°, and side 8 cm opposite 75° — find side *a* (opposite 45°)

Given:
- Angle A = 45°
- Angle B = 75°
- Side b = 8 cm (opposite 75°)
- Find side a (opposite 45°)

Step 1: Find the third angle
Sum of angles in a triangle = 180°
So, angle C = 180° – 45° – 75° = 60°

Step 2: Use Law of Sines
$$
\frac{a}{\sin(45^\circ)} = \frac{8}{\sin(75^\circ)}
$$

Calculate:
- $\sin(45^\circ) \approx 0.7071$
- $\sin(75^\circ) \approx 0.9659$

$$
a = \frac{8 \times \sin(45^\circ)}{\sin(75^\circ)} = \frac{8 \times 0.7071}{0.9659} \approx \frac{5.6568}{0.9659} \approx 5.86 \text{ cm}
$$

✔ Answer: a ≈ 5.86 cm

---

2. Triangle with sides 7 cm and 4 cm, included angle 85° — find side *b*

This is SAS (Side-Angle-Side), so use Law of Cosines.

Given:
- a = 7 cm
- c = 4 cm
- ∠B = 85° (included between a and c)
- Find side b

Law of Cosines:
$$
b^2 = a^2 + c^2 - 2ac \cos(B)
$$
$$
b^2 = 7^2 + 4^2 - 2(7)(4)\cos(85^\circ)
$$
$$
= 49 + 16 - 56 \cos(85^\circ)
$$
- $\cos(85^\circ) \approx 0.0872$

$$
b^2 = 65 - 56(0.0872) = 65 - 4.8832 = 60.1168
$$
$$
b \approx \sqrt{60.1168} \approx 7.75 \text{ cm}
$$

✔ Answer: b ≈ 7.75 cm

---

3. Triangle with angles 105°, 20°, side 6.5 cm — find side *c*

Given:
- Angle A = 20°
- Angle B = 105°
- Side a = 6.5 cm (opposite 20°)
- Find side c (opposite unknown angle)

Step 1: Find third angle
Angle C = 180° – 20° – 105° = 55°

Step 2: Law of Sines
$$
\frac{c}{\sin(55^\circ)} = \frac{6.5}{\sin(20^\circ)}
$$
- $\sin(55^\circ) \approx 0.8192$
- $\sin(20^\circ) \approx 0.3420$

$$
c = \frac{6.5 \times 0.8192}{0.3420} \approx \frac{5.3248}{0.3420} \approx 15.57 \text{ cm}
$$

✔ Answer: c ≈ 15.57 cm

---

4. Triangle with sides 7 cm, 9 cm, included angle 50° — find side *d*

SAS again: two sides and included angle.

Given:
- a = 7 cm
- b = 9 cm
- ∠C = 50° (between them)
- Find side c = d

Law of Cosines:
$$
d^2 = 7^2 + 9^2 - 2(7)(9)\cos(50^\circ)
$$
$$
= 49 + 81 - 126 \cos(50^\circ)
$$
- $\cos(50^\circ) \approx 0.6428$

$$
d^2 = 130 - 126(0.6428) = 130 - 80.8128 = 49.1872
$$
$$
d \approx \sqrt{49.1872} \approx 7.01 \text{ cm}
$$

✔ Answer: d ≈ 7.01 cm

---

5. Triangle with sides 5 cm, 9 cm, angle 33° — find possible value for *e*

Given:
- One side = 5 cm
- Another side = 9 cm
- Angle = 33° — but not specified which angle, but since it’s labeled *e*, likely side e is opposite 33°, and we’re to find a possible value.

Wait — actually, from diagram: angle 33° is at vertex between side 5 cm and side 9 cm? Or not?

Looking at diagram: triangle has:
- Side 5 cm
- Side 9 cm
- Angle 33° adjacent to both? Possibly angle between them → then it’s SAS.

But the instruction says: “find a possible value for *e*”, and *e* is the third side.

So assume:
- Two sides: 5 cm and 9 cm
- Included angle: 33°
- Find side e (opposite the 33° angle)

Wait — no! In standard labeling, side *e* would be opposite angle *E*. But here, if angle 33° is between the two known sides, then side e is the side opposite that angle, so yes, use Law of Cosines.

So:
- a = 5 cm
- b = 9 cm
- ∠C = 33°
- Find side c = e

$$
e^2 = 5^2 + 9^2 - 2(5)(9)\cos(33^\circ)
$$
$$
= 25 + 81 - 90 \cos(33^\circ)
$$
- $\cos(33^\circ) \approx 0.8387$

$$
e^2 = 106 - 90(0.8387) = 106 - 75.483 = 30.517
$$
$$
e \approx \sqrt{30.517} \approx 5.52 \text{ cm}
$$

✔ Answer: e ≈ 5.52 cm

---

6. Triangle with sides 8 cm, 5 cm, 7 cm — find angle *f*

Given:
- Sides: 8 cm, 5 cm, 7 cm
- Find angle f (probably between 8 cm and 5 cm?)

From diagram: side 8 cm and 5 cm meet at angle f, and side 7 cm is opposite.

So angle f is between sides 8 and 5, and opposite side is 7 cm.

Use Law of Cosines:

Let:
- a = 5 cm
- b = 8 cm
- c = 7 cm (opposite angle f)

$$
\cos(f) = \frac{a^2 + b^2 - c^2}{2ab}
= \frac{5^2 + 8^2 - 7^2}{2(5)(8)}
= \frac{25 + 64 - 49}{80}
= \frac{40}{80} = 0.5
$$

$$
f = \cos^{-1}(0.5) = 60^\circ
$$

✔ Answer: f = 60°

---

7. Triangle with sides 4 cm, 7 cm, 6.5 cm — find angle *g*

Given:
- Sides: 4 cm, 7 cm, 6.5 cm
- Angle g is between 4 cm and 7 cm, opposite 6.5 cm?

Wait — from diagram: angle g is between sides 4 cm and 7 cm, and base is 6.5 cm.

So yes: angle g is between 4 and 7, and opposite side is 6.5 cm.

Use Law of Cosines:

Let:
- a = 4 cm
- b = 7 cm
- c = 6.5 cm (opposite angle g)

$$
\cos(g) = \frac{a^2 + b^2 - c^2}{2ab}
= \frac{4^2 + 7^2 - 6.5^2}{2(4)(7)}
= \frac{16 + 49 - 42.25}{56}
= \frac{22.75}{56} \approx 0.40625
$$

$$
g = \cos^{-1}(0.40625) \approx 66.0^\circ
$$

✔ Answer: g ≈ 66.0°

---

8. Triangle with sides 7 cm, 8.5 cm, angle 125° — find side *h*

Given:
- Two sides: 7 cm and 8.5 cm
- Included angle: 125°
- Find side h (opposite the 125° angle)

Use Law of Cosines:

$$
h^2 = 7^2 + 8.5^2 - 2(7)(8.5)\cos(125^\circ)
$$
- $\cos(125^\circ) = \cos(180^\circ - 55^\circ) = -\cos(55^\circ) \approx -0.5736$

$$
h^2 = 49 + 72.25 - 2(7)(8.5)(-0.5736)
= 121.25 + 110.004
= 231.254
$$
$$
h \approx \sqrt{231.254} \approx 15.21 \text{ cm}
$$

✔ Answer: h ≈ 15.21 cm

---

9. Triangle with angles 68°, 42°, side 10 cm — find side *i*

Given:
- Angles: 68°, 42° → third angle = 180° – 68° – 42° = 70°
- Side 10 cm is opposite 68° (from diagram)
- Need to find side i (opposite 42°)

Use Law of Sines:

$$
\frac{i}{\sin(42^\circ)} = \frac{10}{\sin(68^\circ)}
$$
- $\sin(42^\circ) \approx 0.6691$
- $\sin(68^\circ) \approx 0.9272$

$$
i = \frac{10 \times 0.6691}{0.9272} \approx \frac{6.691}{0.9272} \approx 7.22 \text{ cm}
$$

✔ Answer: i ≈ 7.22 cm

---

✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| a | ≈ 5.86 cm |
| b | ≈ 7.75 cm |
| c | ≈ 15.57 cm |
| d | ≈ 7.01 cm |
| e | ≈ 5.52 cm |
| f | 60° |
| g | ≈ 66.0° |
| h | ≈ 15.21 cm |
| i | ≈ 7.22 cm |

---

🔍 Notes on Construction:

To construct these triangles:
- Use a protractor and ruler.
- For ASA or AAS: draw one side, mark angles at endpoints.
- For SAS: draw one side, construct angle, mark second side length.
- For SSS: draw one side, use compass to arc from each end with other lengths.
- Then measure the required side or angle.

These values are approximate due to rounding trigonometric values.

Let me know if you'd like diagrams or steps for construction!