The solution for an oblique triangle can be done with the application of the sine rule and the cosine rule, also known as the Law of Sine and the Law of Cosine. An oblique triangle, as we all know, is a triangle with no right angle. It is a triangle whose angles are all acute or a triangle with one obtuse angle.
The two general forms of an oblique triangle are as shown:
An oblique triangle with all acute angles. Arcs mark the three angles; the dashed line is the perpendicular from the top vertex, meeting the base inside the triangle at the right-angle square.An oblique triangle with one obtuse angle, marked by the arc at the middle vertex. The dashed lines extend the base and drop a perpendicular from the top vertex; the perpendicular meets the extended base outside the triangle, at the right-angle square. A dashed line also retraces the slanted side from the bottom-left vertex to the top vertex, over the solid side.
Sine Rule (The Law of Sine)
The sine rule is used in the following cases:
CASE 1: Given two angles and one side (AAS or ASA)
CASE 2: Given two sides and a non-included angle (SSA)
The sine rule states that the sides of a triangle are proportional to the sines of the opposite angles. In symbols,
In this case, there may be two triangles, one triangle, or no triangle with the given properties. For this reason, it is sometimes called the ambiguous case. Thus, we need to examine the possibility of no solution, one or two solutions.
Below is a summary of these possible solutions of a triangle:
1. If \(\angle A\) is an acute angle and \(a \lt b\), there are three possibilities.
(a) Angle A is acute; side b runs from A up to C. The perpendicular distance from C down to the base line is b sin A (dotted). Side a, swung from C (the short arc), is shorter than b sin A and cannot reach the base line: no triangle is formed.
\(a \lt b \sin A\): no solution.
(b) Angle A is acute; side a, swung from C, just reaches the base line at the single point B, where a = b sin A: exactly one (right-angled) triangle is formed.
\(a = b \sin A\): one solution.
(c) Angle A is acute and b > a > b sin A. The arc of radius a centred at C cuts the base line twice, at the two points labelled B; the dashed segment CB and the solid segment CB are the two possible positions of side a. The dotted perpendicular from C, of length b sin A, meets the base at the right-angle square: two triangles are formed.
\(b \gt a \gt b \sin A\): two solutions.
2. If \(\angle A\) is an acute angle and \(a \geq b\), then there is exactly one solution.
Angle A is acute and a ≥ b. Side a, swung from C (the arc at B), cuts the ray from A exactly once, at B: one triangle ABC, with sides a, b and c.
3. If \(\angle A\) is an obtuse or right angle, there are two possibilities.
Angle A is obtuse. Side b runs from A up to C; side a, swung from C (the short arc), is no longer than b and cannot reach the ray from A towards B: no triangle is formed.
No solution: \(a \leq b\).
Angle A is obtuse. Side a, swung from C, is longer than b and cuts the ray from A exactly once, at B: one triangle ABC, with sides a, b and c, is formed.
One solution: \(a \gt b\).
Cosine Rule (The Law of Cosine)
The cosine rule is used in the following cases:
1. Given two sides and an included angle (SAS)
2. Given three sides (SSS)
The cosine rule states that the square of the length of any side of a triangle equals the sum of the squares of the lengths of the other sides minus twice their product multiplied by the cosine of their included angle. In symbols:
\[ a^2 = b^2 + c^2 - 2bc\cos A \]
\[ b^2 = a^2 + c^2 - 2ac\cos B \]
\[ c^2 = a^2 + b^2 - 2ab\cos C \]
Rearranged to solve for the cosine of each angle:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Go to the next page to start practising what you have learnt.
