Topic 19 Sine and cosine rules

Figure 19.1: Triangle ABC

Theorem 19.1 Given an arbitrary triangle (marked as on Figure 19.1), we have:

1. Sine Rule: \(\boxed{\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}}\) = diameter of the circumscribed circle.

2. Cosine Rule: \(\boxed{a^2+b^2 = c^2 + 2ab\cos C}\)
   and variants:
   \(b^2+c^2 = a^2 + 2bc\cos A\)
   \(a^2+c^2 = b^2 + 2ac\cos B\)

3. \(\boxed{\mathrm{Area} = \frac12 bh = \frac12ab\sin C = \sqrt{s(s-a)(s-b)(s-c)}}\)     where \(h\) is the height of the triangle (see Figure 19.2) and \(s=\dfrac{a+b+c}{2}\).
   (The formula with \(s\) is called Heron’s formula.)

Proofs (non-examinable): Start by further marking the triangle as in Figure 19.2.

1. (Sine rule) By definition of sine in the triangles \(CHB\) and \(HAB\), we have \(a\sin C=h=c\sin A\).
   \(\implies\) \(\dfrac{a}{\sin A}=\dfrac{c}{\sin C}\).
   The other equalities can be derived similarly.

Figure 19.2: Triangle ABC with markings

2. (Cosine rule) We begin by using Pythagoras’ Theorem for the triangles \(CHB\) and \(HAB\): \[\begin{align*} a^2-x^2 &= h^2 = c^2-(b-x)^2\\ a^2-x^2 &=c^2-b^2+2bx - x^2\\ a^2+b^2 &= c^2+2bx \end{align*}\] From the definition of cosine for the triangle \(CHB\), we have \(x=a\cos C\), which leads to \[ a^2+b^2=c^2-2ab\cos C. \]

3. (Area of the triangle)

The area of the triangle \(ABC\) is the sum of the areas of the two right triangles \(CHB\) and \(HAB\), each of which is a half of the area of a rectangle. Thus \[ \mathrm{Area} = \frac12xh + \frac12(b-x)h = \frac12bh. \] Using the definition of sine for the triangle \(CHB\), we get \(h=a\sin C\), hence we get further \[ \mathrm{Area} = \frac12ab\sin C. \] Deriving Heron’s formula is a little more algebraically involved. Recall from above that \[\begin{align*} a^2+b^2&=c^2+2bx\\ \implies\quad\quad\quad x&=\frac{a^2+b^2-c^2}{2b} \end{align*}\] Starting now with the Pythagoras’s Theorem for triangle \(CHB\), we obtain \[\begin{align*} h^2 &= a^2-x^2 = a^2-\left(\frac{a^2+b^2-c^2}{2b}\right)^2\\ &= \frac{4a^2b^2 - (a^2+b^2-c^2)^2}{4b^2}\\ &= \frac{(2ab + a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{4b^2} &&\text{NB: }X^2-Y^2=(X+Y)(X-Y)\\ &= \frac{\left((a+b)^2-c^2\right)\cdot\left(c^2 - (a-b)^2\right)}{4b^2}\\ &= \frac{(a+b+c)(a+b-c)(c+a-b)(c-a+b)}{4b^2} &&\text{NB: same trick as above}\\ &= \frac{2s\cdot 2(s-c)\cdot 2(s-b) \cdot 2(s-a)}{4b^2} = 4\cdot\frac{s(s-a)(s-b)(s-c)}{b^2}. \end{align*}\] Consequently, \[\begin{align*} \mathrm{Area} &= \frac12bh\\ &= \frac12 b \sqrt{4\frac{s(s-a)(s-b)(s-c)}{b^2}}\\ &= \sqrt{s(s-a)(s-b)(s-c)}. \end{align*}\]

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To find all the sides and angles of a triangle, you may want to use the following approach:

given:	use:
one side and two angles	sine rule
two sides and an angle (not between them)	sine rule
two sides and the angle between them	cosine rule
three sides	cosine rule

Example 19.1 Calculate the angles of a triangle with sides of length \(a=80\,\mathrm{mm}\), \(b=70\,\mathrm{mm}\), \(c=50\,\mathrm{mm}\).

Solution. Using the cosine rule, we calculate \[\begin{align*} \cos A &= \frac{b^2+c^2-a^2}{2bc}\\ &= \frac{70^2+50^2-80^2}{2\cdot70\cdot50} =\frac{1}{7} \approx 0.1429 \end{align*}\] \(\implies\) \(\angle A = \arccos 0.1429 \approx 81.78^\circ\).

Using the sine rule, we calculate \[\begin{align*} \sin B &= b\cdot \frac{\sin A}{a}\\ &= 70 \cdot \frac{\sin 81.78^\circ}{80} \approx 0.866 \end{align*}\] \(\implies\) \(\angle B=\arcsin 0.866 \approx 60^\circ\)
\(\implies\) \(\angle C = 180^\circ-\angle A-\angle B= 38.22^\circ\).

Example 19.2 Determine the area of a triangle with \(a=6\,\mathrm{m}\), \(b=5\,\mathrm{m}\), \(c=9\,\mathrm{m}\).

Solution. We use Heron’s formula: \[\begin{align*} s&=\frac{a+b+c+}{2}=\frac{6\,\mathrm{m} + 5\,\mathrm{m}+ 9\,\mathrm{m}}{2}=10\,\mathrm{m}\\ \implies\quad\quad\quad\mathrm{Area} &= \sqrt{s(s-a)(s-b)(s-c)}\\ &= \sqrt{10\,\mathrm{m}\cdot 4\,\mathrm{m}\cdot 5\,\mathrm{m}\cdot 1\,\mathrm{m}} \\ &= \sqrt{200\,\mathrm{m^4}} \approx 14.14\,\mathrm{m^2}. \end{align*}\]

Example 19.3 Determine all sides, angles and area of a triangle with \(b=10\,\mathrm{m}\), \(c=5\,\mathrm{m}\) and \(\angle A=120^\circ\).

Solution. First, we use cosine rule to determine \(a\): \[\begin{align*} a^2 &= b^2+c^2-2bc\cos A\\ &= 100\,\mathrm{m^2} + 25 \,\mathrm{m^2} - 2\cdot 10\,\mathrm{m}\cdot 5\,\mathrm{m}\cdot \cos 120^\circ\\ &= 125\,\mathrm{m^2} + 50 \,\mathrm{m^2} = 175 \,\mathrm{m^2}\\ \implies\quad\quad\quad a &= \sqrt{175\,\mathrm{m^2}} =5\sqrt{7}\,\mathrm{m} \approx 13.23\,\mathrm{m}. \end{align*}\]

Second, we use sine rule to calculate \(\angle B\): \[\begin{align*} \sin B &= b\cdot\frac{\sin A}{a}\\ &= 10\,\mathrm{m}\cdot \frac{\sin 120^\circ}{5\sqrt{7}\,\mathrm{m}} = \frac{10\cdot \frac{\sqrt{3}}{2}}{5\sqrt{7}} =\frac{\sqrt{3}}{\sqrt{7}}\approx 0.6547\\ \implies\quad\quad\quad \angle B &= \arcsin 0.6547 \approx 40.9^\circ. \end{align*}\]

Third, we calculate \[ \angle C = 180^\circ - \angle A-\angle B = 180^\circ- 120^\circ - 40.9^\circ = 19.1^\circ. \]

Finally, we can use (for example) Heron’s formula to obtain the area: \[\begin{align*} s &= \frac{a+b+c}{2} = \frac{10+5+5\sqrt{7}}{2}\,\mathrm{m} \approx 14.12\,\mathrm{m}\\ \implies\quad\quad\quad \mathrm{Area} &= \sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{14.12\cdot 0.89\cdot 4.12\cdot 9.12} \,\mathrm{m^2} \approx 21.73\,\mathrm{m^2}. \end{align*}\]