THE GEOMETRY OF POLYGONS

The base angles are opposite the two equal (marked) sides, and angles opposite equal sides in a triangle are always equal.

$$\frac{180-48}{2}=66° \text{ each}$$

$$x=180-72-72=36°$$

Apex 20°: base angles $=\frac{180-20}{2}=80°$ each. Apex 60°: base angles $=\frac{180-60}{2}=60°$ each. Apex 100°: base angles $=\frac{180-100}{2}=40°$ each.

As the apex angle increases, the base angles decrease (and vice versa) — they change in opposite directions.

This is an EQUILATERAL triangle. It's not a coincidence — an equilateral triangle is actually a special case of an isosceles triangle where ALL sides (not just two) are equal, which happens precisely when the apex angle is 60°, making all angles equal too.

The student found the combined total of BOTH base angles together (130°) but forgot to divide by 2 to find EACH individual base angle (since they are equal, sharing the 130° between them). Correct: $130\div2=65°$ each.

$50+65+65=180°$ ? (while the student's version, $50+130+130=310°$, is clearly wrong since it exceeds 180°).

$$180-65-65=50°$$

New base angles $=\frac{180-34}{2}=73°$ each. Since the apex angle decreased (50°?34°), the tent becomes POINTIER (narrower and taller) at the top.

$$\frac{180-52}{2}=64° \text{ each}$$

$$180-2(63)=54°$$

$$\frac{180-90}{2}=45° \text{ each}$$

If one base angle is 90°, the OTHER base angle must ALSO be 90° (since base angles are equal) — but two 90° angles already sum to 180°, leaving 0° for the apex angle, which is impossible (an angle can't be 0° in a real triangle). So a right angle CANNOT be a base angle of an isosceles triangle; it can only be the apex angle, giving the unique 45-45-90 triangle found in part (a).

Apex 20°: base $=\frac{160}{2}=80°$ each. Apex 80°: base $=\frac{100}{2}=50°$ each. Apex 140°: base $=\frac{40}{2}=20°$ each.

Base angle $=\frac{180-\text{apex}}{2}$. When apex=20°, base=80°. When apex=140°, base=20° — the SPECIFIC swap occurs here because $140=180-2(20)$, a special numerical coincidence in this case (not a general rule for all isosceles triangles) — but it usefully illustrates that apex and base angles are linked through the fixed 180° total, so extreme values of one naturally correspond to extreme (but different) values of the other.

The student found the COMBINED total of both base angles (142°) but didn't divide by 2 to find each INDIVIDUAL angle. Correct: $142\div2=71°$ each.

If each base angle were 142°, the two base angles ALONE would sum to $142\times2=284°$ — already exceeding the maximum possible total of 180° for an entire triangle, before even including the apex angle. Any single angle in a valid triangle must be less than 180°, and certainly two of them can't sum to more than 180° on their own — this alone signals an error.

This claim is false in general — in a 'wide, flat' isosceles triangle (large apex angle, small base angles), the BASE (the non-equal side) can actually be LONGER than the two equal sides. For example, with a very large apex angle close to 180°, the triangle becomes almost flat, with the base stretching out to be much longer than the two short equal sides. The relative length of the equal sides versus the base depends entirely on the SPECIFIC apex angle, not a fixed rule.

$$\frac{180-36}{2}=72°\text{ each}$$

For equilateral (all angles equal), each angle must be $180\div3=60°$. So the apex angle would need to change from 36° to 60°, and ALL three angles (apex and both base angles) would then equal 60° each.

$$180-2(55)=70°$$

Since base angle $=\frac{180-\text{apex}}{2}$, a LARGER apex angle directly causes a SMALLER base angle (they're inversely related through this fixed formula) — so achieving the LARGEST possible apex angle while still meeting the 'base angle ?55°' requirement means using the SMALLEST allowed base angle exactly at its minimum (55°), which is precisely what gives the maximum apex angle of 70°; any larger apex angle would push the base angle below the required 55° minimum.