Non-Right-Angled Triangle Trigonometry

Consider triangle \(ABC\) with base \(a = BC\). Drop a perpendicular height \(h\) from vertex \(A\) to the base \(a\), meeting \(BC\) at point \(H\). The area of a triangle is given by the standard formula: $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}ah. $$ In the right-angled triangle \(AHC\), the sine ratio gives: $$ \sin(C) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{b} \implies h = b\sin(C). $$ Substituting this expression for \(h\) into the area formula gives: $$ \text{Area} = \frac{1}{2}a(b\sin C) = \frac{1}{2}ab\sin C. $$ (The figure places the foot of the altitude inside the triangle — the acute case; if the angle at \(C\) is obtuse, the foot \(H\) falls outside, beyond \(C\); the right-angled triangle \(AHC\) then has the angle \(180^\circ - C\) at \(C\), so the height is \(h = b\sin(180^\circ - C) = b\sin C\): the formula is unchanged.)

Drop a perpendicular height \(h\) from vertex \(A\) to the side \(a\) (line \(\Line{BC}\)), meeting it at point \(D\). In the right-angled triangle \(ADC\):

1. By Pythagoras' theorem, \(h^2 + x^2 = b^2 \implies h^2 = b^2 - x^2\).
2. \(\cos(C) = x/b \implies x = b\cos(C)\).

In the right-angled triangle \(ADB\), by Pythagoras' theorem: $$ c^2 = h^2 + (a-x)^2. $$ Substitute the expressions for \(h^2\) and \(x\) into this equation: $$ \begin{aligned} c^2 &= (b^2 - x^2) + (a^2 - 2ax + x^2) \\ &= a^2 + b^2 - 2ax \\ &= a^2 + b^2 - 2a(b\cos C) \\ &= a^2 + b^2 - 2ab\cos C. \end{aligned} $$ The figure places the foot \(D\) on the side \(\Segment{BC}\), between \(B\) and \(C\): this is the case when the angles \(B\) and \(C\) are both acute, even if the angle \(A\) is obtuse. If the angle \(B\) or the angle \(C\) is obtuse, the foot \(D\) falls outside the side \(\Segment{BC}\). For \(B\) obtuse, \(D\) lies beyond \(B\): \(BD = x - a\) and \(BD^2 = (a - x)^2\), so the computation above is unchanged. For \(C\) obtuse, \(D\) lies beyond \(C\), and the right-angled triangle \(ADC\) has the angle \(180^\circ - C\) at \(C\): \(CD = b\cos(180^\circ - C) = -b\cos C\), \(BD = a + CD = a - b\cos C\) and \(h^2 = b^2 - CD^2 = b^2 - b^2\cos^2 C\). Then $$\begin{aligned}[t] c^2 &= h^2 + BD^2 && {\scriptstyle(\text{Pythagoras' theorem in } ADB)}\\ &= b^2 - b^2\cos^2 C + a^2 - 2ab\cos C + b^2\cos^2 C\\ &= a^2 + b^2 - 2ab\cos C, \end{aligned}$$ the same formula.

The area of the triangle can be expressed in three different ways using the area formula: $$ \text{Area} = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B = \frac{1}{2} ab \sin C. $$ Dividing each part of the equality by \(\frac{1}{2}abc\) gives: $$ \frac{bc \sin A}{abc} = \frac{ac \sin B}{abc}= \frac{ab \sin C}{abc}. $$ So $$ \frac{\sin A}{a} = \frac{\sin B}{b}= \frac{\sin C}{c}. $$ Taking the reciprocal of each term gives the Law of Sines: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. $$

A triangle gives an admissible angle. Suppose that a triangle with the angle \(A\) and the sides \(a\) (opposite \(A\)) and \(b\) exists, and call \(B\) its angle opposite \(b\). The Law of Sines gives \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\), that is \(\sin B = \dfrac{b \sin A}{a}\), and the angle sum of the triangle gives \(A + B < 180^\circ\). An admissible angle gives a triangle. Conversely, let \(B\) be an angle with \(0^\circ < B < 180^\circ\), \(\sin B = \dfrac{b \sin A}{a}\) and \(A + B < 180^\circ\). Then \(C = 180^\circ - A - B\) is a positive angle, and there is a triangle, unique up to congruence, with a side \(\Segment{BC}\) of length \(a\) and the angles \(B\) and \(C\) at its ends (the case “a side and the two angles at its ends”); its third angle is \(180^\circ - B - C = A\). By the Law of Sines, its side opposite \(B\) measures $$\begin{aligned}[t] CA &= \frac{a \sin B}{\sin A} && \text{(Law of Sines)}\\ &= \frac{a}{\sin A} \times \frac{b \sin A}{a} && \text{(value of } \sin B\text{)}\\ &= b. \end{aligned}$$ So the triangles with the data \(A\), \(a\), \(b\) correspond exactly to the angles \(B\) with \(\sin B = \dfrac{b \sin A}{a}\) and \(A + B < 180^\circ\): one triangle for each such angle. It remains to count these angles, with the number \(s = \dfrac{b \sin A}{a} > 0\).

• If \(s > 1\): the sine of an angle is at most \(1\), so no angle \(B\) exists — no triangle.
• If \(s = 1\): on the unit circle, the only point of height \(1\) is the point for \(90^\circ\), so \(B = 90^\circ\) is the only candidate. It passes the test \(A + 90^\circ < 180^\circ\) exactly when \(A\) is acute: then exactly one triangle, right-angled at \(B\); otherwise none.
• If \(s < 1\): the horizontal line of height \(s\) cuts the upper half of the unit circle at two points, symmetric about the vertical axis, so the angles between \(0^\circ\) and \(180^\circ\) whose sine is \(s\) are the acute angle \(B_1 = \sin^{-1}(s)\) and its supplement \(B_2 = 180^\circ - B_1\) (indeed \(\sin(180^\circ - B_1) = \sin B_1\)). Each of them gives a triangle exactly when it passes the test \(A + B < 180^\circ\): zero, one or two triangles.