THE GEOMETRY OF POLYGONS

$$x=180-55-70=55°$$

$$x=180-90-35=55°$$

The other two angles must each be acute (less than 90°), since they must sum to exactly 90° together, and neither can be 0° or more than 90°.

$60+60+60=180°$ ?. $40+50+90=180°$ ?. Both confirm the angle sum rule.

Since the torn corners always seem to form a straight line (180°) regardless of the triangle's shape, this suggests the property holds universally, not just for special triangles — though a fully rigorous mathematical proof (e.g. using parallel line angle facts) is needed to confirm it holds for absolutely every possible triangle.

Third angle $=180-100-95=-15°$ — a NEGATIVE angle, which is impossible for a real triangle.

If two angles were both obtuse (each greater than 90°), their sum alone would already exceed 180° — leaving no room (or even a negative amount) for the third angle. Since all three angles must be positive and sum to exactly 180°, at most ONE angle in any triangle can be obtuse.

$48 + 2y + y = 180 \Rightarrow 3y=132 \Rightarrow y=44°$. So the second angle is $2(44)=88°$.

Angles: $48°, 88°, 44°$. Since the largest angle (88°) is less than 90°, this is an ACUTE triangle.

$$x=180-65-48=67°$$

$x=180-90-27=63°$. Since the triangle contains a 90° angle (and no angle exceeds 90°), it is classified as a RIGHT triangle.

Exterior angle $=180-70=110°$.

$55+55=110°$ — EXACTLY matches the exterior angle found. This confirms the exterior angle theorem: the exterior angle of a triangle always equals the sum of the two interior angles that are NOT adjacent to it.

$$A+B+C=180°$$

From $A+B+C=180°$: $C=180-A-B$. Substituting into $E+C=180°$: $E+(180-A-B)=180 \Rightarrow E=A+B$. This proves the exterior angle theorem holds for ANY triangle, not just specific tested examples — it follows directly and necessarily from the two fundamental facts (angle sum of a triangle, and supplementary angles on a straight line).

Since $E=A+B$, and BOTH $A$ and $B$ must be positive angles (every angle in a triangle is greater than 0°), adding a positive amount to either one INDIVIDUALLY must always make $E$ LARGER than that angle alone — i.e. $E=A+B>A$ (since $B>0$) and similarly $E=A+B>B$ (since $A>0$). This isn't a coincidence limited to specific examples; it's a guaranteed consequence of the exterior angle theorem itself, true for every possible triangle.

The student found the SUM of the two given angles but incorrectly used this sum AS the third angle, instead of subtracting it from 180° (the full angle sum of a triangle). Correct: $180-47-68=65°$.

If the third angle were genuinely 115°, the total sum would be $47+68+115=230°$, far exceeding the required 180° angle sum for ANY triangle — recognizing that all three angles must sum to EXACTLY 180° (not more) is a quick way to catch this kind of error before even doing the correct calculation.

$38+(3y+15)+y=180 \Rightarrow 4y+53=180 \Rightarrow 4y=127 \Rightarrow y=31.75°$.

Angles: $38°$, $3(31.75)+15=110.25°$, $31.75°$. Since the largest angle (110.25°) exceeds 90°, this is an OBTUSE triangle.

$$180-90-6=84°$$

$\tan(6°)=\frac{0.9}{\text{run}} \Rightarrow \text{run}=\frac{0.9}{\tan(6°)}\approx8.6$m. A STEEPER angle (8.5° instead of 6°) would require a SHORTER horizontal run to achieve the same 0.9m rise, since a steeper incline gains height more quickly over a shorter horizontal distance — so choosing 8.5° would result in a shorter (but steeper, less gentle) ramp compared to using 6°.