Solve triangles from sides or an included angle

Try

Two straight paths of lengths 5 m and 7 m leave a common point at an included angle of 60°. Find the distance between their endpoints. The Pythagorean theorem alone is not applicable because the included angle is not right. The cosine law gives c²=25+49−70cos60°=39, so c=√39 m.

Predict

Keep the two path lengths fixed and increase the included angle to 120°. Cosine becomes negative, so the subtraction term increases c². Predict why the endpoints move farther apart. At 90°, the cosine term vanishes and the familiar right-triangle formula returns.

Experiment

Place the first endpoint at (a,0) and the second at (b cos C,b sin C). Draw their joining segment, opposite angle C. Use equal scales on both axes so that the picture preserves geometric angles and distances.

Included C	cos C	c² for a=5,b=7
60°	1/2	39
90°	0	74
120°	−1/2	109

The coordinate-distance expression is (a−b cos C)²+(b sin C)². Expand it and combine the two b² terms using sin²C+cos²C=1. This derives, rather than merely states, the cosine law.

Understand

For a triangle with side c opposite included angle C, c²=a²+b²−2ab cos C. Given three sides, rearrange to cos C=(a²+b²−c²)/(2ab), then use arccos to recover the unique triangle angle in (0,π). Check positive sides and strict triangle inequalities first.

If c² exceeds a²+b², cosine is negative and C is obtuse. Equality gives a right angle; a smaller c² gives an acute angle. To classify the whole triangle from its largest angle, use its longest side as c.

The area is ab sin C/2. This follows from base a times altitude b sin C, divided by two. It works for obtuse included angles too: the altitude may lie outside the drawn side segment, but its positive length is still b sin C. The angle must be the one between the two side lengths used.

Computable inverse values do not replace feasibility checks. Sides 2, 3 and 5 yield a cosine of −1 for the largest angle, but their vertices are collinear and area is zero. They are a degenerate boundary, not a nondegenerate triangle. Sides 2, 3 and 6 fail even the non-strict inequality and produce an impossible cosine outside the real inverse range.

Apply

The practice includes squared lengths so exact irrational side lengths do not need decimal checking. In the derivation, label which side is opposite the angle. In the invalid-data task, explain the geometric failure rather than treating it as a calculator error.

Continue

The coordinate derivation contains a dot product in disguise. Vectors will generalize the relationship between lengths, angles and projected components without requiring a triangle diagram for every calculation.