Law of Sines worksheet with problems for finding missing sides and angles in triangles, designed for geometry and trigonometry practice.

To solve the problems in the worksheet, we will use the Law of Sines. The Law of Sines states:

\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]

where \(a\), \(b\), and \(c\) are the sides of the triangle, and \(A\), \(B\), and \(C\) are the angles opposite those sides, respectively.

Section A: Find the missing sides

#### 1)
Given:
- \(\angle B = 29^\circ\)
- \(\angle C = 82^\circ\)
- Side \(b = ?\)

First, find \(\angle A\):
\[
\angle A = 180^\circ - \angle B - \angle C = 180^\circ - 29^\circ - 82^\circ = 69^\circ
\]

Using the Law of Sines:
\[
\frac{b}{\sin 29^\circ} = \frac{18}{\sin 69^\circ}
\]

Solve for \(b\):
\[
b = \frac{18 \cdot \sin 29^\circ}{\sin 69^\circ}
\]

Calculate:
\[
\sin 29^\circ \approx 0.4848, \quad \sin 69^\circ \approx 0.9336
\]
\[
b = \frac{18 \cdot 0.4848}{0.9336} \approx \frac{8.7264}{0.9336} \approx 9.35
\]

So, \(b \approx 9.35\) cm.

#### 2)
Given:
- \(\angle A = 74^\circ\)
- \(\angle C = 33^\circ\)
- Side \(a = 9\) cm
- Side \(c = ?\)

First, find \(\angle B\):
\[
\angle B = 180^\circ - \angle A - \angle C = 180^\circ - 74^\circ - 33^\circ = 73^\circ
\]

Using the Law of Sines:
\[
\frac{a}{\sin A} = \frac{c}{\sin C}
\]
\[
\frac{9}{\sin 74^\circ} = \frac{c}{\sin 33^\circ}
\]

Solve for \(c\):
\[
c = \frac{9 \cdot \sin 33^\circ}{\sin 74^\circ}
\]

Calculate:
\[
\sin 33^\circ \approx 0.5446, \quad \sin 74^\circ \approx 0.9613
\]
\[
c = \frac{9 \cdot 0.5446}{0.9613} \approx \frac{4.9014}{0.9613} \approx 5.10
\]

So, \(c \approx 5.10\) cm.

#### 3)
Given:
- \(\angle A = 68^\circ\)
- \(\angle C = 40^\circ\)
- Side \(b = 15.7\) cm
- Side \(c = ?\)

First, find \(\angle B\):
\[
\angle B = 180^\circ - \angle A - \angle C = 180^\circ - 68^\circ - 40^\circ = 72^\circ
\]

Using the Law of Sines:
\[
\frac{b}{\sin B} = \frac{c}{\sin C}
\]
\[
\frac{15.7}{\sin 72^\circ} = \frac{c}{\sin 40^\circ}
\]

Solve for \(c\):
\[
c = \frac{15.7 \cdot \sin 40^\circ}{\sin 72^\circ}
\]

Calculate:
\[
\sin 40^\circ \approx 0.6428, \quad \sin 72^\circ \approx 0.9511
\]
\[
c = \frac{15.7 \cdot 0.6428}{0.9511} \approx \frac{10.12676}{0.9511} \approx 10.65
\]

So, \(c \approx 10.65\) cm.

Section B: Find the missing angles

#### 1)
Given:
- Side \(XY = 8\) cm
- Side \(YZ = 11\) cm
- \(\angle Z = 30^\circ\)
- \(\angle X = ?\)

Using the Law of Sines:
\[
\frac{XY}{\sin Z} = \frac{YZ}{\sin X}
\]
\[
\frac{8}{\sin 30^\circ} = \frac{11}{\sin X}
\]

Solve for \(\sin X\):
\[
\sin X = \frac{11 \cdot \sin 30^\circ}{8}
\]

Calculate:
\[
\sin 30^\circ = 0.5
\]
\[
\sin X = \frac{11 \cdot 0.5}{8} = \frac{5.5}{8} = 0.6875
\]

Find \(\angle X\):
\[
X = \sin^{-1}(0.6875) \approx 43.4^\circ
\]

So, \(\angle X \approx 43.4^\circ\).

#### 2)
Given:
- Side \(XY = 23\) cm
- Side \(YZ = 14\) cm
- \(\angle X = 27^\circ\)
- \(\angle Z = ?\)

Using the Law of Sines:
\[
\frac{XY}{\sin Z} = \frac{YZ}{\sin X}
\]
\[
\frac{23}{\sin Z} = \frac{14}{\sin 27^\circ}
\]

Solve for \(\sin Z\):
\[
\sin Z = \frac{23 \cdot \sin 27^\circ}{14}
\]

Calculate:
\[
\sin 27^\circ \approx 0.4540
\]
\[
\sin Z = \frac{23 \cdot 0.4540}{14} = \frac{10.442}{14} \approx 0.7459
\]

Find \(\angle Z\):
\[
Z = \sin^{-1}(0.7459) \approx 48.3^\circ
\]

So, \(\angle Z \approx 48.3^\circ\).

#### 3)
Given:
- Side \(XZ = 108\) mm
- Side \(YZ = 110\) mm
- \(\angle X = 62^\circ\)
- \(\angle Y = ?\)

Using the Law of Sines:
\[
\frac{XZ}{\sin Y} = \frac{YZ}{\sin X}
\]
\[
\frac{108}{\sin Y} = \frac{110}{\sin 62^\circ}
\]

Solve for \(\sin Y\):
\[
\sin Y = \frac{108 \cdot \sin 62^\circ}{110}
\]

Calculate:
\[
\sin 62^\circ \approx 0.8829
\]
\[
\sin Y = \frac{108 \cdot 0.8829}{110} = \frac{95.3052}{110} \approx 0.8664
\]

Find \(\angle Y\):
\[
Y = \sin^{-1}(0.8664) \approx 60.1^\circ
\]

So, \(\angle Y \approx 60.1^\circ\).

Section C: Find the missing side

#### 1)
Given:
- \(\angle P = 69^\circ\)
- \(\angle Q = 79^\circ\)
- Side \(PR = 18\) cm
- Side \(PQ = x\)

First, find \(\angle R\):
\[
\angle R = 180^\circ - \angle P - \angle Q = 180^\circ - 69^\circ - 79^\circ = 32^\circ
\]

Using the Law of Sines:
\[
\frac{PR}{\sin Q} = \frac{PQ}{\sin R}
\]
\[
\frac{18}{\sin 79^\circ} = \frac{x}{\sin 32^\circ}
\]

Solve for \(x\):
\[
x = \frac{18 \cdot \sin 32^\circ}{\sin 79^\circ}
\]

Calculate:
\[
\sin 79^\circ \approx 0.9816, \quad \sin 32^\circ \approx 0.5299
\]
\[
x = \frac{18 \cdot 0.5299}{0.9816} \approx \frac{9.5382}{0.9816} \approx 9.72
\]

So, \(x \approx 9.72\) cm.

#### 2)
Given:
- \(\angle A = 75^\circ\)
- \(\angle B = 48^\circ\)
- Side \(AC = 12\) cm
- Side \(BC = b\)

First, find \(\angle C\):
\[
\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 75^\circ - 48^\circ = 57^\circ
\]

Using the Law of Sines:
\[
\frac{AC}{\sin B} = \frac{BC}{\sin A}
\]
\[
\frac{12}{\sin 48^\circ} = \frac{b}{\sin 75^\circ}
\]

Solve for \(b\):
\[
b = \frac{12 \cdot \sin 75^\circ}{\sin 48^\circ}
\]

Calculate:
\[
\sin 75^\circ \approx 0.9659, \quad \sin 48^\circ \approx 0.7431
\]
\[
b = \frac{12 \cdot 0.9659}{0.7431} \approx \frac{11.5908}{0.7431} \approx 15.60
\]

So, \(b \approx 15.60\) cm.

#### 3)
Given:
- \(\angle Y = 29^\circ\)
- \(\angle Z = 112^\circ\)
- Side \(YZ = 8\) cm
- Side \(XY = z\)

First, find \(\angle X\):
\[
\angle X = 180^\circ - \angle Y - \angle Z = 180^\circ - 29^\circ - 112^\circ = 39^\circ
\]

Using the Law of Sines:
\[
\frac{YZ}{\sin X} = \frac{XY}{\sin Z}
\]
\[
\frac{8}{\sin 39^\circ} = \frac{z}{\sin 112^\circ}
\]

Solve for \(z\):
\[
z = \frac{8 \cdot \sin 112^\circ}{\sin 39^\circ}
\]

Calculate:
\[
\sin 112^\circ \approx 0.9272, \quad \sin 39^\circ \approx 0.6293
\]
\[
z = \frac{8 \cdot 0.9272}{0.6293} \approx \frac{7.4176}{0.6293} \approx 11.80
\]

So, \(z \approx 11.80\) cm.

Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{Section A:} & b \approx 9.35 \text{ cm}, \quad a \approx 5.10 \text{ cm}, \quad c \approx 10.65 \text{ cm} \\
\text{Section B:} & x \approx 43.4^\circ, \quad z \approx 48.3^\circ, \quad y \approx 60.1^\circ \\
\text{Section C:} & x \approx 9.72 \text{ cm}, \quad b \approx 15.60 \text{ cm}, \quad z \approx 11.80 \text{ cm}
\end{array}
}
\]