Quadrilateral from sides and angles

What counts as "included." An included angle sits between two of the given sides — it is the angle actually enclosed by them, not any angle that happens to touch that vertex. If $AB$, $BC$, $CD$ are given, the natural included angles are $\angle B$ (between $AB$ and $BC$) and $\angle C$ (between $BC$ and $CD$). Mixing this up — using an exterior angle, or an angle at a vertex that is not between the two named sides — derails the whole construction before the first arc is drawn.

Why the walk works. Fixing one side fixes two points. Turning through a known angle at one endpoint fixes a direction; marking a known length along that direction fixes the next point. Repeating this once more locks a third point. Three points already position the whole figure in the plane (up to where you choose to start), so the fourth vertex is not free — it is simply wherever the last two given sides happen to land, and the final side is whatever straight segment joins them. This is why there is no ambiguity of the kind seen in some diagonal-based constructions, where an arc can cross a line at two different points and give two different quadrilaterals: here, each step produces exactly one new point.

where: each arrow is one construction move — draw a side of the stated length from the current vertex, then at the new vertex swing the protractor through the stated included angle to fix the direction of the next side. Repeating the chain three times (side, angle, side, angle, side) places all four vertices; joining the last vertex back to the first closes the quadrilateral.

Three sides and the two angles between them determine the quadrilateral completely, with no second, differently shaped figure possible from the same measurements — because every vertex after the first two is placed by exactly one length and exactly one angle, leaving nothing to choose.

Step 1. Fix a starting side. Draw $AB$ to the given length. This pins down two points, $A$ and $B$, and with them a fixed direction (the line $AB$) from which every later angle will be measured.

Step 2. Fix the next direction. At $B$, the given included angle $\angle B$ is the angle between $AB$ and $BC$. Constructing that exact angle with a protractor gives one, and only one, ray along which $BC$ can run — swing the protractor to any other reading and the angle drawn would not match the data.

Step 3. Fix the next point. Marking the known length $BC$ along that single ray gives exactly one point $C$; there is no second candidate, unlike constructions that intersect two arcs (which can meet at two symmetric points). The same two moves — construct the given angle, then mark the given length — repeat at $C$ to fix $D$ using $\angle C$ and $CD$.

Step 4. Close the figure. Once $A$, $B$, $C$, $D$ are all located, the last side is not new data — it is simply the straight segment $DA$ joining the last point back to the first. Because every step from $A$ onward involved exactly one forced choice, the entire quadrilateral, including the two "leftover" angles $\angle A$ and $\angle D$, is completely determined by the original three sides and two angles; nothing about the figure is left for the student to decide.

Construct quadrilateral $ABCD$ with $AB = 5.0\ \text{cm}$, $BC = 4.5\ \text{cm}$, $CD = 4.0\ \text{cm}$, $\angle B = 75^\circ$, and $\angle C = 110^\circ$.

Step 1. Sketch a rough plan first, marking the given lengths and angles roughly at the correct vertices, so the real construction is not attempted blind. Then draw $AB = 5.0\ \text{cm}$ accurately with a ruler.

Step 2. At $B$, use a protractor to construct $\angle ABX = 75^\circ$, where ray $BX$ shows the direction of side $BC$. Along $BX$, mark off $BC = 4.5\ \text{cm}$ with the compass or ruler to fix point $C$.

Step 3. At $C$, construct $\angle BCY = 110^\circ$ with the protractor, reading it on the same side as the rough sketch, giving ray $CY$ for the direction of $CD$. Along $CY$, mark $CD = 4.0\ \text{cm}$ to fix point $D$.

Step 4. Join $D$ back to $A$ with a straight line. $ABCD$, with $AB = 5.0\ \text{cm}$, $BC = 4.5\ \text{cm}$, $CD = 4.0\ \text{cm}$, $\angle B = 75^\circ$, $\angle C = 110^\circ$, and closing side $DA$, is the required quadrilateral — a simple, non-crossed figure whose remaining angles $\angle A$ and $\angle D$ can be checked with a protractor but were never independently chosen.

Competitive stretch

Using $PQ = 4.2\ \text{cm}$, $QR = 5.2\ \text{cm}$, $RS = 3.8\ \text{cm}$, $\angle Q = 100^\circ$, $\angle R = 95^\circ$, the identical four-step walk (draw $PQ$; at $Q$ turn $100^\circ$ and mark $QR$; at $R$ turn $95^\circ$, measured consistently on the same side, and mark $RS$; join $S$ back to $P$) produces quadrilateral $PQRS$. Here too, the closing side $SP$ and both remaining angles $\angle P$, $\angle S$ come out of the drawing automatically — a useful check that the method generalises to any labelling, not just $ABCD$.

• Measuring the included angle on the wrong side of the line, which flips the figure or creates a self-intersecting (crossed) quadrilateral. Keep the rough sketch beside the real drawing and check which side each new ray belongs on.
• Confusing "included angle" with any angle at that vertex — it must be the specific angle between the two named sides, not an exterior angle or an angle formed with a diagonal.
• Marking a length from the wrong point, such as measuring $CD$ from $B$ instead of from $C$. Every new length is always marked from the vertex most recently located.
• Treating the closing side as a free choice. The last side ($DA$ or $SP$) is not given data — it is whatever segment results from joining the final point back to the first, and must not be redrawn to a "nicer" length.

Ask the tutor: not sure which side of the line to swing the protractor on for your own numbers, or want a step-by-step walkthrough of this construction — ask the AI tutor to work through it with you vertex by vertex.