Trigonometry (A) Missing Lengths worksheet with 12 problems involving right triangles and angle measurements.

Let's solve each of the trigonometry problems in Section A and Section B, where we are to find the missing side lengths using trigonometric ratios (SOH CAH TOA), given an angle and one side.

We'll use:
- SOH: $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$
- CAH: $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$
- TOA: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$

All answers should be given to 3 significant figures.

---

✔ Section A

#### 1) Right triangle:
- Angle = $36^\circ$, opposite side = $x$, hypotenuse = $12$ cm
Use SOH:
$$
\sin(36^\circ) = \frac{x}{12} \Rightarrow x = 12 \cdot \sin(36^\circ)
$$
$$
x = 12 \cdot 0.5878 \approx 7.0536 \Rightarrow \boxed{7.05} \text{ cm}
$$

---

#### 2)
- Angle = $18^\circ$, adjacent = $9$ cm, opposite = $x$
Use TOA:
$$
\tan(18^\circ) = \frac{x}{9} \Rightarrow x = 9 \cdot \tan(18^\circ)
$$
$$
x = 9 \cdot 0.3249 \approx 2.9241 \Rightarrow \boxed{2.92} \text{ cm}
$$

---

#### 3)
- Angle = $49^\circ$, adjacent = $18$ cm, opposite = $x$
Use TOA:
$$
\tan(49^\circ) = \frac{x}{18} \Rightarrow x = 18 \cdot \tan(49^\circ)
$$
$$
x = 18 \cdot 1.1504 \approx 20.7072 \Rightarrow \boxed{20.7} \text{ cm}
$$

---

#### 4)
- Angle = $62^\circ$, adjacent = $17$ cm, hypotenuse = $x$
Use CAH:
$$
\cos(62^\circ) = \frac{17}{x} \Rightarrow x = \frac{17}{\cos(62^\circ)}
$$
$$
x = \frac{17}{0.4695} \approx 36.25 \Rightarrow \boxed{36.3} \text{ cm}
$$

---

#### 5)
- Angle = $27^\circ$, opposite = $5$ cm, hypotenuse = $x$
Use SOH:
$$
\sin(27^\circ) = \frac{5}{x} \Rightarrow x = \frac{5}{\sin(27^\circ)}
$$
$$
x = \frac{5}{0.4540} \approx 11.01 \Rightarrow \boxed{11.0} \text{ cm}
$$

---

#### 6)
- Angle = $32^\circ$, opposite = $21$ cm, hypotenuse = $x$
Use SOH:
$$
\sin(32^\circ) = \frac{21}{x} \Rightarrow x = \frac{21}{\sin(32^\circ)}
$$
$$
x = \frac{21}{0.5299} \approx 39.64 \Rightarrow \boxed{39.6} \text{ cm}
$$

---

✔ Section B

#### 1)
- Angle = $18^\circ$, adjacent = $14$ cm, hypotenuse = $x$
Use CAH:
$$
\cos(18^\circ) = \frac{14}{x} \Rightarrow x = \frac{14}{\cos(18^\circ)}
$$
$$
x = \frac{14}{0.9511} \approx 14.72 \Rightarrow \boxed{14.7} \text{ cm}
$$

---

#### 2)
- Angle = $71^\circ$, opposite = $19$ cm, hypotenuse = $x$
Use SOH:
$$
\sin(71^\circ) = \frac{19}{x} \Rightarrow x = \frac{19}{\sin(71^\circ)}
$$
$$
x = \frac{19}{0.9455} \approx 20.09 \Rightarrow \boxed{20.1} \text{ cm}
$$

---

#### 3)
- Angle = $42^\circ$, adjacent = $5$ cm, hypotenuse = $x$
Use CAH:
$$
\cos(42^\circ) = \frac{5}{x} \Rightarrow x = \frac{5}{\cos(42^\circ)}
$$
$$
x = \frac{5}{0.7431} \approx 6.726 \Rightarrow \boxed{6.73} \text{ cm}
$$

---

#### 4)
- Angle = $34^\circ$, opposite = $11$ cm, hypotenuse = $x$
Use SOH:
$$
\sin(34^\circ) = \frac{11}{x} \Rightarrow x = \frac{11}{\sin(34^\circ)}
$$
$$
x = \frac{11}{0.5592} \approx 19.66 \Rightarrow \boxed{19.7} \text{ cm}
$$

---

#### 5)
- Angle = $70^\circ$, adjacent = $7$ cm, opposite = $x$
Use TOA:
$$
\tan(70^\circ) = \frac{x}{7} \Rightarrow x = 7 \cdot \tan(70^\circ)
$$
$$
x = 7 \cdot 2.7475 \approx 19.2325 \Rightarrow \boxed{19.2} \text{ cm}
$$

---

#### 6)
- Angle = $36^\circ$, adjacent = $24$ cm, opposite = $x$
Use TOA:
$$
\tan(36^\circ) = \frac{x}{24} \Rightarrow x = 24 \cdot \tan(36^\circ)
$$
$$
x = 24 \cdot 0.7265 \approx 17.436 \Rightarrow \boxed{17.4} \text{ cm}
$$

---

✔ Final Answers:

#### Section A
1) $x = \boxed{7.05}$ cm
2) $x = \boxed{2.92}$ cm
3) $x = \boxed{20.7}$ cm
4) $x = \boxed{36.3}$ cm
5) $x = \boxed{11.0}$ cm
6) $x = \boxed{39.6}$ cm

#### Section B
1) $x = \boxed{14.7}$ cm
2) $x = \boxed{20.1}$ cm
3) $x = \boxed{6.73}$ cm
4) $x = \boxed{19.7}$ cm
5) $x = \boxed{19.2}$ cm
6) $x = \boxed{17.4}$ cm

---

✔ All answers are rounded to 3 significant figures as requested. Let me know if you'd like a diagram or explanation for any specific problem!