Derive triangle laws and measure inaccessible distances

Try

Two sides of a triangle are b=6 cm and c=5 cm, with included angle A=120 degrees. The triangle is obtuse, so an altitude may meet an extended side rather than the side segment. Predict whether the familiar area formula involving sine still works.

Use the conventional labels a=BC, b=CA and c=AB, with angle A opposite a. Drop the perpendicular from C to the line AB. Its positive height is b sin A=6 sin120°=3√3 cm. The area is c times height divided by two, so it is 15√3/2 cm². The perpendicular foot lies beyond A; this does not make the height negative.

Predict

For an acute angle A, the foot falls along the forward ray from A, and the right triangle gives h=b sin A directly. If A is obtuse, the small right triangle uses the acute supplementary angle 180°−A. Unit-circle symmetry gives sin(180°−A)=sin A, so the same positive expression holds. At A=90° the altitude coincides with AC and h=b.

Predict what would fail if we used b cos A as the height. Cosine describes the horizontal projection, which is negative in the obtuse case. Height is the perpendicular distance to the baseline, not the signed position of the perpendicular foot.

Experiment

Construct an obtuse triangle from A=(0,0), B=(6,0), C=(−2,4). Drop the perpendicular to the full line AB and label D=(−2,0). The picture is a separate coordinate example, not the 120-degree triangle above.

Now move C to (2,4), keeping A and B. The height and area remain the same, but the triangle becomes acute at A. Explain which quantity the move preserves and why equal area does not imply congruence.

Understand

For any nondegenerate triangle, h=b sin A and also h=a sin B when measured from C to line AB. If one of A or B is obtuse, the supplementary-angle argument supplies the same identity; both cannot be obtuse. Dividing the two positive expressions gives b sin A=a sin B, hence a/sin A=b/sin B.

Repeat the argument using the altitude from B to line AC. It gives c sin A=a sin C, so a/sin A=c/sin C. Together these establish the sine law

a/sin A=b/sin B=c/sin C.

The sines are positive because each interior angle lies strictly between 0 and 180 degrees. The proof treats exterior perpendicular feet explicitly and does not assume that every altitude stays inside the triangle.

The same heights give area K=bc sin A/2. Cyclic relabeling yields K=ca sin B/2=ab sin C/2. For an obtuse configuration, the elementary base-times-height area rule can itself be seen by subtracting two right triangles that share an external altitude: the baseline differences leave the original base length. No signed negative area is needed.

As a use of the sine law, let A=30°, B=45° and a=6. Then b=6 sin45°/sin30°=6√2, and the third angle is 105°. The law relates opposite pairs; swapping b with the angle adjacent to it would invalidate the proportion.

When a law gives only a sine value, two angles may share it: θ and 180°−θ. Check both candidates against the remaining angle sum, side data and nondegeneracy. An inverse-sine calculator returns one principal value, not an automatic guarantee of a unique triangle. The existing SSA ambiguity lesson supplies the full classification.

Apply

Draw and label an altitude before deriving an area or sine-law relation. A complete derivation names the base, positive perpendicular height, corresponding opposite side/angle pairs and any supplementary angle. For a measurement problem, choose a law from the supplied data instead of guessing from the diagram.

Record degree or radian mode for approximate calculations and round only after the main calculation. Reject negative angles, angle sums of 180 degrees before the third positive angle is included, or side lengths that violate triangle inequality. The correct algebraic output can still describe a degenerate or incompatible configuration.

Continue

Reuse the cosine-law derivation when two sides and their included angle, or three sides, are known. Revisit SSA ambiguity before accepting a single inverse-sine result.

Optional depth: the common sine-law ratio is twice the circumradius. That requires a separate circle-angle argument; it is not assumed in the altitude proof above. Keeping the proof routes separate makes their dependencies inspectable.