Question

1. calculate the measure of each lettered angle.

Explanation:

Step1: Identify the polygon types

The left figure is a regular hexagon (all sides and angles equal), the top is a square (all sides and angles equal, right angle \(90^\circ\)), the right is a regular pentagon (all sides and angles equal). The angles \(a\), \(b\), \(c\), \(d\) are around a point (sum to \(360^\circ\)), and \(e\) is in a triangle.

Step2: Calculate angle \(a\) (hexagon interior angle)

For a regular hexagon, the formula for interior angle is \(\frac{(n - 2)\times180^\circ}{n}\), where \(n = 6\).
\(\frac{(6 - 2)\times180^\circ}{6}=\frac{4\times180^\circ}{6}=120^\circ\). So \(a = 120^\circ\).

Step3: Calculate angle \(b\) (pentagon interior angle)

For a regular pentagon, \(n = 5\). Interior angle: \(\frac{(5 - 2)\times180^\circ}{5}=\frac{3\times180^\circ}{5}=108^\circ\). So \(b = 108^\circ\).

Step4: Calculate angle \(c\) (square interior angle)

Square has interior angle \(90^\circ\), so \(c = 90^\circ\).

Step5: Calculate angle \(d\) (around a point)

Angles around a point sum to \(360^\circ\). So \(d=360^\circ - a - b - c=360^\circ - 120^\circ - 108^\circ - 90^\circ = 42^\circ\)? Wait, no—wait, the triangle: the triangle has two equal sides (marked), so it's isosceles? Wait, no, the lines from the center: wait, maybe I misread. Wait, the angles \(a\), \(b\), \(c\), \(d\) are around a point, but also, the triangle: let's re - examine.

Wait, the square has angle \(90^\circ\) (so \(c = 90^\circ\)), hexagon has \(120^\circ\) (\(a = 120^\circ\)), pentagon has \(108^\circ\) (\(b = 108^\circ\)). Then \(d=360-(120 + 108+90)=360 - 318 = 42^\circ\)? No, wait, maybe the triangle: the triangle is formed with the square and pentagon? Wait, no, the triangle has two sides equal (marked with one tick), so it's isosceles. Wait, the angle \(e\): in the triangle, the angle at \(d\) and the other angle? Wait, maybe I made a mistake. Let's correct:

Wait, the regular hexagon interior angle: \(\frac{(6 - 2)\times180}{6}=120^\circ\) (correct). Regular square: \(90^\circ\) (correct). Regular pentagon: \(\frac{(5 - 2)\times180}{5}=108^\circ\) (correct). Now, around the central point, the angles \(a\) (hexagon), \(b\) (pentagon), \(c\) (square), and \(d\) (from the triangle's adjacent angle) should sum to \(360^\circ\)? Wait, no, the lines: the square, hexagon, pentagon, and the triangle are all meeting at the center. So \(a + b + c + d=360^\circ\). So \(d = 360-(120 + 108+90)=42^\circ\). Then the triangle: it has two sides equal (marked with one tick), so it's isosceles. The angle \(d\) is \(42^\circ\), so the other two angles (including \(e\)): wait, no, the triangle's angles: the triangle is formed with the square's side and pentagon's side? Wait, no, the triangle has two sides equal (so two angles equal). Wait, the angle at \(d\) is \(42^\circ\), so the triangle's angles: \(180 - 42=138^\circ\), divided by 2? No, wait, maybe the triangle is equilateral? No, the ticks: one tick on two sides. Wait, maybe I messed up the angle \(d\). Wait, let's re - express:

Wait, the regular hexagon: each interior angle \(120^\circ\), so the angle at the center for the hexagon's "corner" is \(120^\circ\) ( \(a = 120^\circ\) ). The square: interior angle \(90^\circ\) ( \(c = 90^\circ\) ). The pentagon: interior angle \(108^\circ\) ( \(b = 108^\circ\) ). Then \(d=360-(120 + 90+108)=42^\circ\). Then the triangle: it has two sides equal (marked with one tick), so it's isosceles with vertex angle \(d = 42^\circ\)? No, wait, the triangle's sides: the lines from the center to the vertices—maybe the triangle is formed by the square and the other figure. Wait, no, the triangle has two sides with one tick, so it's isosceles, so the base angles are equal. So \(e=\frac{180 - d}{2}=\frac{180 - 42}{2}=69^\circ\)? No, this seems wrong. Wait, maybe the initial assumption of the polygons is wrong.

Wait, another approach: the regular hexagon has central angle (angle at center between two vertices) \(60^\circ\) (since \(360/6 = 60^\circ\)), but interior angle is \(120^\circ\). The square has central angle \(90^\circ\) (interior angle \(90^\circ\)). The pentagon has central angle \(72^\circ\) ( \(360/5 = 72^\circ\) ), but interior angle \(108^\circ\). Wait, I think I confused interior angle with the angle around the center.

Wait, the angle \(a\) is the interior angle of the hexagon, which is \(120^\circ\) (correct, because in a regular hexagon, interior angle \(= 180 - 60=120^\circ\), since central angle is \(60^\circ\)). The angle \(b\) is the interior angle of the pentagon, \(108^\circ\) (correct, \(180 - 72 = 108^\circ\), central angle \(72^\circ\)). The angle \(c\) is the interior angle of the square, \(90^\circ\) (correct, \(180 - 90 = 90^\circ\), central angle \(90^\circ\)). Then, around the common vertex, the sum of angles \(a + b + c + d=360^\circ\), so \(d = 360-(120 + 108+90)=42^\circ\). Then the triangle: it has two sides equal (marked with one tick), so it's isosceles, so the two base angles are equal. So \(e=\frac{180 - d}{2}=\frac{180 - 42}{2}=69^\circ\)? No, this doesn't seem right. Wait, maybe the triangle is equilateral? No, the ticks are one tick, not three. Wait, maybe I made a mistake in the polygon types.

Wait, let's start over:

1. Regular Hexagon:
Number of sides \(n = 6\).
Interior angle formula: \(I=\frac{(n - 2)\times180^\circ}{n}\).
\(I=\frac{(6 - 2)\times180^\circ}{6}=120^\circ\). So \(a = 120^\circ\).

2. Square (regular quadrilateral, \(n = 4\)):
Interior angle \(I=\frac{(4 - 2)\times180^\circ}{4}=90^\circ\). So \(c = 90^\circ\).

3. Regular Pentagon (\(n = 5\)):
Interior angle \(I=\frac{(5 - 2)\times180^\circ}{5}=108^\circ\). So \(b = 108^\circ\).

4. Angle \(d\) (around a point, sum to \(360^\circ\)):
\(d=360^\circ - a - b - c=360 - 120 - 108 - 90 = 42^\circ\).

5. Triangle: The triangle has two sides equal (marked with one tick), so it's isosceles. The angle at \(d\) is \(42^\circ\), so the other two angles (including \(e\)) are equal.
\(e=\frac{180^\circ - d}{2}=\frac{180 - 42}{2}=69^\circ\)? Wait, no, maybe the triangle is not with vertex angle \(d\). Wait, maybe the triangle is formed by the square and the hexagon? No, the square has angle \(90^\circ\), the hexagon \(120^\circ\), the pentagon \(108^\circ\).

Wait, maybe the angles \(a\), \(b\), \(c\), \(d\) are the angles between the sides of the polygons at the common vertex. So:

- \(a\): angle from hexagon: \(120^\circ\)
- \(b\): angle from pentagon: \(108^\circ\)
- \(c\): angle from square: \(90^\circ\)
- \(d\): \(360-(120 + 108+90)=42^\circ\)

Then the triangle: it has two sides equal (marked), so it's isosceles with base angles equal. So \(e=\frac{180 - d}{2}=69^\circ\)? But this seems odd. Wait, maybe I made a mistake in the polygon identification. Let's check the tick marks:

- Hexagon: all angles with one tick, so regular hexagon (all angles equal).
- Square: all angles with two ticks, regular square (all angles equal, \(90^\circ\)).
- Pentagon: all angles with three ticks, regular pentagon (all angles equal, \(108^\circ\)).
- Triangle: two sides with one tick, so isosceles.

So summarizing:

- \(a = 120^\circ\) (hexagon interior angle)
- \(b = 108^\circ\) (pentagon interior angle)
- \(c = 90^\circ\) (square interior angle)
- \(d=360 - 120 - 108 - 90 = 42^\circ\)
- \(e=\frac{180 - 42}{2}=69^\circ\) (isosceles triangle, base angles equal)

Answer:

\(a = 120^\circ\), \(b = 108^\circ\), \(c = 90^\circ\), \(d = 42^\circ\), \(e = 69^\circ\)