Trigonometry/Law of Sines

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Law-of-sines1.svg

For any triangle with vertices A, B, and C, corresponding angles A, B, and C, and corresponding opposite side lengths a, b, and c, the Law of Sines states that

{a \over \sin A} = {b \over \sin B} = {c \over \sin C}.


Each of these expressions is also equal to the diameter of the triangle's circumcircle (the circle that passes through the points A, B, and C). The law can also be written in terms of the reciprocals:

{\sin A \over a} = {\sin B \over b} = {\sin C \over c}.

Proof[edit]

Law-of-sines2.svg

Dropping a perpendicular OC from vertex C to intersect AB (or AB extended) at O splits this triangle into two right-angled triangles AOC and BOC. We can calculate the length h of the altitude OC in two different ways:

  • Using the triangle AOC gives
\displaystyle h=b \sin A ;
  • and using the triangle BOC gives
\displaystyle h=a \sin B .
  • Eliminate h from these two equations:
\displaystyle a \sin B =b \sin A .
  • Rearrange to obtain
\displaystyle {a \over \sin A} = {b \over \sin B}.

By using the other two perpendiculars the full law of sines can be proved. QED.

Application[edit]

This formula can be used to find the other two sides of a triangle when one side and the three angles are known. (If two angles are known, the third is easily found since the sum of the angles is 180º.) See Solving Triangles Given ASA. It can also be used to find an angle when two sides and the angle opposite one side are known.

Area of a triangle[edit]

The area of a triangle may be found in various ways. If all three sides are known, use Heron's theorem.

If two sides and the included angle are known, consider the second diagram above. Let the sides b and c, and the angle between them α be known. From triangle ACO, the altitude h = CO is bsin(α) so the area is 12cbsin(α).

If two angles and the included side are known, again consider the second diagram above. Let the side c and the angles α and γ be known. Let AO =x. Then

\displaystyle \frac{x}{h} = \cot(\alpha) \text {; } \frac{c-x}{h} = \cot(\gamma) \text {; adding these, } \frac{c}{h} = \cot(\alpha) + \cot(\gamma)

Thus

\displaystyle h = \frac{c}{\cot(\alpha) + \cot(\gamma)} \text { so  area } = \frac{c^2}{2(\cot(\alpha) + \cot(\gamma))}.