Construction of Triangles

A closed triangle cannot be formed from an arbitrary set of three segment lengths. Two strict conditions must be fulfilled:

1. Identify the longest side among the three given lengths.
2. Calculate the sum of the two shorter side lengths.
3. If the sum is strictly greater ($>$) than the longest side, a unique triangle can be constructed. If the sum is less than ($<$) or equal to ($=$), the arcs will never intersect in a point above the line.

⚠️ Equality Trap ($a + b = c$): When $a + b = c$ (e.g., $3\text{ cm}, 4\text{ cm}, 7\text{ cm}$), the arcs meet directly on the base line segment, producing a degenerate flat line with zero area, not a triangle!

Structural Truss Engineering: Civil engineers designing Warren and Pratt bridge trusses rely on triangle inequality to ensure load-bearing struts never fail or undergo buckling under dynamic railway loads.

Under the Side-Side-Side (SSS) criterion, specifying three valid side lengths uniquely fixes the triangle up to congruence. No protractor is required; the geometric intersection of two compass circles uniquely pinpoints the third vertex.

1. Rough Sketch: Sketch a rough triangle with labeled vertices $A, B, C$ and side lengths.
2. Base Construction: Draw a straight ray and use the compass to mark off base $BC = a$.
3. First Arc: With $B$ as center and radius equal to side $c$, draw an arc above $BC$.
4. Second Arc: With $C$ as center and radius equal to side $b$, draw an arc intersecting the first arc at point $A$.
5. Join Vertices: Connect $A$ to $B$ and $A$ to $C$ using a straightedge. The required $\triangle ABC$ is constructed.

Problem: Construct $\triangle ABC$ where $BC = 6.5\text{ cm}, AB = 5.2\text{ cm}$, and $AC = 4.8\text{ cm}$.
Procedure: Lay off $BC = 6.5\text{ cm}$. From $B$, swing an arc of radius $5.2\text{ cm}$. From $C$, swing an arc of radius $4.8\text{ cm}$. Label the point of intersection $A$ and join $AB$ and $AC$.

Compass Radius Drift: Always check compass hinge tightness before swinging arcs. A minor $2\text{ mm}$ shift distorts the side lengths.

GPS Satellite Trilateration: Global Positioning System receivers compute precise user latitude, longitude, and elevation by intersecting sphere distances from GPS satellites using the SSS principle.

In SAS (Side-Angle-Side), the given angle MUST be the strictly included angle lying directly between the two given arms. If a non-included angle is provided (SSA/ASS), an ambiguous case arises yielding either two triangles, one triangle, or none.

1. Draw base line segment $QR$ of given length.
2. At vertex $Q$, construct the specified angle (e.g., $60^\circ$) using compass to form ray $QX$.
3. From ray $QX$, cut off segment $QP$ equal to the second given side length.
4. Join vertex $P$ to vertex $R$. The unique triangle $\triangle PQR$ is established.

Problem: Construct $\triangle ABC$ with $BC = 6\text{ cm}, AB = 4.5\text{ cm}$, and included $\angle B = 60^\circ$.
Solution: Draw $BC = 6\text{ cm}$. Construct $60^\circ$ at $B$. Cut $4.5\text{ cm}$ along the angular arm to locate $A$. Join $AC$. Measuring gives $AC \approx 5.5\text{ cm}$.

Constructing the angle at vertex $C$ instead of vertex $B$: Always verify which vertex the angle letter designates before drawing.

Robotic Arm Forward Kinematics: In industrial pick-and-place robots, the lengths of the upper arm and forearm along with the elbow joint angle determine the precise gripper coordinates via SAS geometry.

When a side and its two adjacent angles are known, it is direct ASA. When two angles and a non-adjacent side are given (AAS), compute the third angle via $\angle 3 = 180^\circ - (\angle 1 + \angle 2)$ and construct on the base line segment via ASA.

1. Draw given base segment $BC$.
2. Construct the first base angle at $B$ and extend its ray.
3. Construct the second base angle at $C$ with its arm sloping inwards.
4. The intersection point of these two angular rays is vertex $A$. $\triangle ABC$ is complete.

Problem: Construct $\triangle XYZ$ where $YZ = 5.5\text{ cm}, \angle Y = 60^\circ$, and $\angle X = 75^\circ$.
Calculation: $\angle Z = 180^\circ - (60^\circ + 75^\circ) = 45^\circ$.
Construction: Draw $YZ = 5.5\text{ cm}$. Construct $60^\circ$ at $Y$ and $45^\circ$ at $Z$. Rays intersect at $X$, automatically forming $\angle X = 75^\circ$.

Angle Sum Exceeding $180^\circ$: If the sum of two given angles is $\ge 180^\circ$, the rays diverge or run parallel, making triangle construction impossible.

Coastal Vessel Triangulation: Navigators fix a ship's position at sea by observing bearings (angles) from two charted lighthouses separated by a known baseline distance.

In RHS (Right-angle, Hypotenuse, Side), the angle is implicitly fixed at $90^\circ$. The side opposite the right angle is the hypotenuse ($h$). By Pythagoras' theorem: $h^2 = a^2 + b^2$, meaning the hypotenuse must always be the strictly longest side ($h > a$).

1. Draw given leg as base segment $BC$.
2. At $B$, erect a perpendicular ($90^\circ$) ray $BY$ using compass.
3. With center at $C$ and radius equal to the given hypotenuse length, swing an arc cutting ray $BY$ at $A$.
4. Join $A$ to $C$. Right-angled triangle $\triangle ABC$ is constructed.

Problem: Construct a right-angled $\triangle ABC$ where $\angle B = 90^\circ$, base $BC = 4\text{ cm}$, and hypotenuse $AC = 5\text{ cm}$.
Verification: Measuring altitude $AB$ yields exactly $3\text{ cm}$, confirming $\sqrt{5^2 - 4^2} = \sqrt{9} = 3\text{ cm}$.

Swinging Hypotenuse Arc from the Wrong Vertex: Never swing the hypotenuse arc from vertex $B$ (where the $90^\circ$ angle is). It MUST be swung from the opposite baseline end $C$ to intersect the vertical ray!

Carpentry & Masonry 3-4-5 Squaring: Construction builders align corner walls to perfect $90^\circ$ angles using RHS geometry with lengths in ratio 3:4:5.