Right triangles are easy: SOH-CAH-TOA does the job. But most triangles you meet in surveying, navigation and engineering have no right angle. This chapter gives you two powerful tools, the Law of Sines and the Law of Cosines, that let you solve any triangle, find its area, and answer questions about bearings and distances.
1. Oblique triangles and the naming convention
An oblique triangle is a triangle with no right angle. To solve a triangle means to find all three sides and all three angles.
To solve one, you need three well-chosen measurements. Throughout the chapter we use one convention: side \(a\) is opposite angle \(A\), side \(b\) is opposite angle \(B\), and side \(c\) is opposite angle \(C\). Angles add up to \(180^\circ\), and the longest side always faces the largest angle.
The possible “given” patterns are named by the order in which the known parts appear around the triangle: AAS or ASA (two angles and a side), SAS (two sides and the angle between them), SSS (three sides) and SSA (two sides and an angle not between them). Each pattern points to a different strategy, as you will see in Part 5.
2. The Law of Sines
In every triangle \(ABC\), \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}.\] This common value equals \(2R\), where \(R\) is the radius of the circle passing through the three vertices.
Why it works. Draw the altitude \(h\) from \(C\) to side \(AB\). It gives \(h=b\sin A\) and also \(h=a\sin B\). So \(b\sin A=a\sin B\), which rearranges to \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\). Repeating with another altitude brings in \(c\).
In triangle \(ABC\), \(A=40^\circ\), \(B=65^\circ\) and \(a=12\) cm. Find \(C\), \(b\) and \(c\).
First, \(C=180^\circ-40^\circ-65^\circ=75^\circ\). Then \(b=\dfrac{a\sin B}{\sin A}=\dfrac{12\sin 65^\circ}{\sin 40^\circ}\approx 16.92\text{ cm}\) and \(c=\dfrac{12\sin 75^\circ}{\sin 40^\circ}\approx 18.03\text{ cm}\).
3. The ambiguous case (SSA)
When you know two sides and an angle that is not between them, the Law of Sines gives \(\sin B\), and an angle with a given sine can be acute or obtuse: \(B\) and \(180^\circ-B\) share the same sine. Sometimes both choices produce a real triangle, sometimes only one, sometimes none.
- Compute \(\sin B=\dfrac{b\sin A}{a}\). If the result is greater than 1, no triangle exists.
- If \(\sin B\le 1\), find the acute angle \(B_1=\sin^{-1}(\sin B)\) and the obtuse candidate \(B_2=180^\circ-B_1\).
- Keep a candidate only if \(A+B<180^\circ\). Each valid \(B\) gives \(C=180^\circ-A-B\) and then \(c=\dfrac{a\sin C}{\sin A}\).
With \(A\) acute and \(h=b\sin A\), the outcomes are summarized below.
| Condition | Number of triangles |
|---|---|
\(a| None |
|
| \(a=h\) | One (right triangle) |
\(h| Two |
|
| \(a\ge b\) | One |
If \(A\) is right or obtuse, there is exactly one triangle when \(a>b\) and none otherwise.
Let \(A=30^\circ\), \(a=5\) and \(b=8\). Here \(h=8\sin30^\circ=4<5<8\), so expect two triangles. We get \(\sin B=\dfrac{8\sin 30^\circ}{5}=0.8\), so \(B_1\approx 53.13^\circ\) and \(B_2\approx 126.87^\circ\). Both satisfy \(A+B<180^\circ\).
Triangle 1: \(C_1\approx 96.87^\circ\), \(c_1=\dfrac{5\sin 96.87^\circ}{\sin 30^\circ}\approx 9.93\). Triangle 2: \(C_2\approx 23.13^\circ\), \(c_2\approx 3.93\).
4. The Law of Cosines
In every triangle \(ABC\), \[c^2=a^2+b^2-2ab\cos C,\] and likewise \(a^2=b^2+c^2-2bc\cos A\) and \(b^2=a^2+c^2-2ac\cos B\). Solved for an angle: \[\cos C=\dfrac{a^2+b^2-c^2}{2ab}.\]
When \(C=90^\circ\), \(\cos C=0\) and the formula collapses to the Pythagorean theorem, so the Law of Cosines is its generalization. If \(C\) is obtuse, \(\cos C<0\), so \(c^2>a^2+b^2\).
Let \(a=7\), \(b=10\) and \(C=48^\circ\). Then \(c^2=49+100-140\cos 48^\circ\approx 55.32\), so \(c\approx 7.44\). To find \(A\), use the Law of Sines (safe here, because \(a\) is the shortest side so \(A\) is acute): \(\sin A=\dfrac{7\sin 48^\circ}{7.44}\), giving \(A\approx 44.4^\circ\) and \(B\approx 87.6^\circ\).
Sides \(6\), \(8\) and \(11\). Start with the largest angle, opposite \(11\): \(\cos C=\dfrac{36+64-121}{2\cdot 6\cdot 8}=-\dfrac{21}{96}\), so \(C\approx 102.6^\circ\), an obtuse angle. Starting with the largest angle is smart: after that, the other two angles must be acute, so the Law of Sines can finish the job without ambiguity.
5. Choosing the right law
| Given | First step |
|---|---|
| AAS or ASA | Find the third angle, then Law of Sines |
| SAS | Law of Cosines for the third side |
| SSS | Law of Cosines for the largest angle |
| SSA | Law of Sines, then check for 0, 1 or 2 triangles |
Do not use the Law of Sines to find an angle that might be obtuse unless you have checked the alternative \(180^\circ-B\). The cosine formula avoids the problem, because cosine is negative for obtuse angles and distinguishes them.
6. Area of a triangle using sine
The area of a triangle with sides \(a\) and \(b\) enclosing angle \(C\) is \[K=\dfrac12\,ab\sin C.\]
This is just \(\dfrac12\times\text{base}\times\text{height}\) with height \(h=b\sin C\). The same formula works with any pair of sides and the angle between them.
A triangular garden bed has sides 9 ft and 14 ft meeting at \(35^\circ\). Its area is \(K=\dfrac12(9)(14)\sin35^\circ\approx 36.14\text{ ft}^2\), which is about \(3.36\text{ m}^2\).
7. Heron’s formula
When you know all three sides and no angle, you can skip the cosine entirely.
Let \(s=\dfrac{a+b+c}{2}\) be the semiperimeter. Then \[K=\sqrt{s(s-a)(s-b)(s-c)}.\]
For sides 13, 14 and 15: \(s=21\), so \(K=\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84\). Check with the figure: \(\dfrac12(14)(12)=84\).
8. Bearings and navigation
In navigation, a bearing is an angle measured clockwise from north and written with three digits, such as \(065^\circ\) (a little north of east) or \(210^\circ\) (south-southwest). The key to bearing problems: draw a north line at every turning point, then compute the interior angle of the triangle from the bearings.
A ship leaves port P and sails 40 nautical miles (nmi, about 74 km) on bearing \(065^\circ\) to point Q, then 55 nmi on bearing \(150^\circ\) to point R. How far is R from the port?
At Q, the direction back to P has bearing \(065^\circ+180^\circ=245^\circ\). The new heading is \(150^\circ\), so the interior angle at Q is \(245^\circ-150^\circ=95^\circ\). By the Law of Cosines, \(d^2=40^2+55^2-2(40)(55)\cos95^\circ\approx 5008.5\), so \(d\approx 70.8\) nmi (about 131 km).
On my home planet we navigate by two moons, and we always sketch first! A rough north-arrow diagram at each turn catches most sign errors before they happen.
Key takeaways
- Law of Sines: \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\). Use it with AAS, ASA, or SSA.
- SSA can give 0, 1 or 2 triangles: compare \(a\) with \(h=b\sin A\) and with \(b\).
- Law of Cosines: \(c^2=a^2+b^2-2ab\cos C\). Use it with SAS or SSS, and find the largest angle first.
- Area: \(K=\dfrac12ab\sin C\), or Heron’s \(K=\sqrt{s(s-a)(s-b)(s-c)}\) when only sides are known.
- Bearings are measured clockwise from north; draw a north line at every turn.
Test yourself: quick challenge for Grade 12
Speed drill for Grade 12: how many in 60 seconds?
🚀 Keep exploring with Zyro
✏️ Math practiceLaw of Sines and Law of Cosines: math practice, Grade 12
📝 Math testsLaw of Sines and Law of Cosines: math test, Grade 12
🎯 Math quizzesLaw of Sines and Law of Cosines: math quiz, Grade 12
✏️ Math practiceVectors, Polar Coordinates, and Complex Numbers: math practice, Grade 12
✏️ Math practiceConic Sections and Parametric Equations: math practice, Grade 12
📝 Math testsVectors, Polar Coordinates, and Complex Numbers: math test, Grade 12