Law of Sines Worksheet | PDF Printable Geometry Worksheet - Free Printable

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:

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