• Based on triangle \(ABC\) above, the sine rule is given by:

\(\dfrac{a}{\sin{A}}=\dfrac{b}{\sin{B}}=\dfrac{c}{\sin{C}}\)

• Alternatively, it can be written as:

\(\dfrac{\sin{A}}{a}=\dfrac{\sin{B}}{b}=\dfrac{\sin{C}}{c}\)

• Where \(a\), \(b\), and \(c\) are the lengths of the sides opposite angles \(A\), \(B\), and \(C\) respectively.

• When two angles and one side are known.
• When two sides and a non-included angle are known.

• In right-angled triangles for which the Sine Rule can be replaced by basic trigonometric ratios.

• Given two angles and one side, use the sine rule to find the other sides.

• Given two sides and one non-included angle, use the sine rule to find the unknown angle.

• When given two sides and a non-included angle, the triangle may have two solutions (two possible triangles), one solution (one triangle), or no solution.

• When two sides and one non-included angle (\(a\), \(b\), \(\angle{A}\)) are given, an ambiguous case occurs if:
  • \(a \lt b\),
  • \(\angle{A}\) is an acute angle.
• Two triangles exist (triangle \(AB_1C\) and \(AB_2C\)).

In the diagram above,   \(LMN\) is a triangle.

Find the length of \(LN\).

Based on the question,

\(\begin{aligned} \angle{M}&=180^\circ -85^ \circ-40^\circ \\ &=55 ^\circ \end{aligned}\)

Applying the Sine Rule:

\(\begin{aligned} \dfrac{LN}{\sin 55^\circ}&=\dfrac{7}{ \sin 40 ^\circ} \\\\ LN&= \dfrac{7}{ \sin 40 ^\circ} \times \sin 55^\circ \\\\ &= 8.92 \text{ cm} \end{aligned}\)

Given a triangle \(ABC\) such that \(AB=6.2\) cm, \(AC=4.8\) cm, and \(\angle{ABC}=43^\circ\).

Find the possible values of \(\angle{BCA}\).

Apply the Sine Rule to find the possible angles of \(C\):

\(\begin{aligned} \dfrac{\sin C}{6.2}&=\dfrac{\sin 43^\circ}{ 4.8} \\\\ \angle BC_1A&=\sin^{-1} \left( \dfrac{\sin 43^\circ}{4.8} \times 6.2 \right) \\\\ &=61.75 ^ \circ \\\\ \angle BC_2A&= 180 ^\circ-61.75 ^ \circ \\\\ &=118.25 ^\circ \end{aligned}\)

• Based on triangle \(ABC\) above, the sine rule is given by:

\(\dfrac{a}{\sin{A}}=\dfrac{b}{\sin{B}}=\dfrac{c}{\sin{C}}\)

• Alternatively, it can be written as:

\(\dfrac{\sin{A}}{a}=\dfrac{\sin{B}}{b}=\dfrac{\sin{C}}{c}\)

• Where \(a\), \(b\), and \(c\) are the lengths of the sides opposite angles \(A\), \(B\), and \(C\) respectively.

• When two angles and one side are known.
• When two sides and a non-included angle are known.

• In right-angled triangles for which the Sine Rule can be replaced by basic trigonometric ratios.

• Given two angles and one side, use the sine rule to find the other sides.

• Given two sides and one non-included angle, use the sine rule to find the unknown angle.

• When given two sides and a non-included angle, the triangle may have two solutions (two possible triangles), one solution (one triangle), or no solution.

• When two sides and one non-included angle (\(a\), \(b\), \(\angle{A}\)) are given, an ambiguous case occurs if:
  • \(a \lt b\),
  • \(\angle{A}\) is an acute angle.
• Two triangles exist (triangle \(AB_1C\) and \(AB_2C\)).

In the diagram above,   \(LMN\) is a triangle.

Find the length of \(LN\).

Based on the question,

\(\begin{aligned} \angle{M}&=180^\circ -85^ \circ-40^\circ \\ &=55 ^\circ \end{aligned}\)

Applying the Sine Rule:

\(\begin{aligned} \dfrac{LN}{\sin 55^\circ}&=\dfrac{7}{ \sin 40 ^\circ} \\\\ LN&= \dfrac{7}{ \sin 40 ^\circ} \times \sin 55^\circ \\\\ &= 8.92 \text{ cm} \end{aligned}\)

Given a triangle \(ABC\) such that \(AB=6.2\) cm, \(AC=4.8\) cm, and \(\angle{ABC}=43^\circ\).

Find the possible values of \(\angle{BCA}\).

Apply the Sine Rule to find the possible angles of \(C\):

\(\begin{aligned} \dfrac{\sin C}{6.2}&=\dfrac{\sin 43^\circ}{ 4.8} \\\\ \angle BC_1A&=\sin^{-1} \left( \dfrac{\sin 43^\circ}{4.8} \times 6.2 \right) \\\\ &=61.75 ^ \circ \\\\ \angle BC_2A&= 180 ^\circ-61.75 ^ \circ \\\\ &=118.25 ^\circ \end{aligned}\)