Question

1. find the values of the indicated angles

Answer:

Left Figure (Pentagon - Isosceles Triangle and Isosceles Trapezoid)

Step 1: Analyze the Triangle (Top)

The top triangle is isosceles (two equal sides marked). The base of the triangle is equal to the top base of the trapezoid (marked equal). The trapezoid is also isosceles (non - parallel sides marked equal). For a regular pentagon, the interior angle is \(108^{\circ}\), but here we can also use triangle and trapezoid properties.

The triangle at the top: Let's consider the fact that the trapezoid's base angles and the triangle's base angles are related. The triangle is isosceles, and if we consider the fact that the trapezoid has base angles equal. Also, the sum of angles in a triangle is \(180^{\circ}\). If we assume the triangle is equilateral? No, but maybe the triangle is isosceles with the base equal to the side of the trapezoid. Wait, another approach: In a regular pentagon, the interior angle is \(\frac{(5 - 2)\times180^{\circ}}{5}=108^{\circ}\). The trapezoid part: the angle \(d\) is an interior angle of the isosceles trapezoid which is part of the pentagon - like figure. The triangle at the top: the two equal sides of the triangle and the equal sides of the trapezoid. The angle \(c\): since the triangle is isosceles and the trapezoid's base is parallel to the top of the triangle, the triangle is actually an isosceles triangle with the base angles equal. If we consider that the trapezoid's non - parallel sides are equal, and the triangle's sides are equal to the trapezoid's non - parallel sides. The sum of angles in a triangle is \(180^{\circ}\), and if we consider that the angle at the top of the triangle and the two base angles \(c\) (since it's isosceles, the two base angles are equal). Wait, maybe the figure is a combination of an isosceles triangle and an isosceles trapezoid, and the triangle is equilateral? No, let's think again.

Wait, the left figure: the top is an isosceles triangle, and the bottom is an isosceles trapezoid. The sides of the triangle (the equal sides) are equal to the non - parallel sides of the trapezoid. The base of the triangle is equal to the top base of the trapezoid. So the triangle and the trapezoid form a pentagon - like shape. For the angle \(c\): in the isosceles triangle, if we consider that the trapezoid's base angles and the triangle's base angles are such that the triangle is isosceles with the two base angles equal. Also, the interior angle of a regular pentagon is \(108^{\circ}\), and the trapezoid's interior angle \(d = 108^{\circ}\)? Wait, no. Wait, the sum of angles in a triangle is \(180^{\circ}\), and if we consider that the triangle is isosceles and the angle at the top of the triangle and the two angles \(c\) add up to \(180^{\circ}\). Also, the trapezoid: in an isosceles trapezoid, base angles are equal. The angle \(d\) and the adjacent angle (if we extend the sides) would be supplementary, but maybe a better way: the triangle is isosceles with the two equal sides, and the base is equal to the side of the trapezoid. If we consider that the triangle is actually an isosceles triangle with the two base angles \(c = 36^{\circ}\)? Wait, no, let's calculate the interior angle of a regular pentagon: \(\theta=\frac{(n - 2)\times180^{\circ}}{n}\), for \(n = 5\), \(\theta=\frac{3\times180^{\circ}}{5}=108^{\circ}\). The trapezoid's angle \(d\): in the isosceles trapezoid, the base angles are equal. The triangle at the top: the two equal sides, and the base is parallel to the trapezoid's top base. So the triangle is isosceles, and the angle \(c\): since the trapezoid's base angle \(d = 108^{\circ}\), and the triangle's base angle \(c\) and the trapezoid's base angle are related. Wait, the triangle and the trapezoid form a pentagon, so the sum of angles around a point? No, let's use the fact that in the isosceles triangle, the two base angles \(c\) and the vertex angle (top angle of the triangle) sum to \(180^{\circ}\). Also, the trapezoid's base angles \(d\) and the adjacent angles: in an isosceles trapezoid, the base angles are equal, and the sum of a base angle and the adjacent non - base angle (if we consider the triangle) is \(180^{\circ}\)? No, the trapezoid has two parallel sides (the two bases) and two non - parallel sides (equal). The sum of interior angles of a trapezoid is \(360^{\circ}\), so if it's isosceles, the two base angles are equal, and the two non - base angles are equal. Let the base angles be \(d\) and the non - base angles be \(x\). Then \(2d + 2x=360^{\circ}\), so \(d + x = 180^{\circ}\). The triangle at the top: it's isosceles, with the two equal sides equal to the non - parallel sides of the trapezoid (\(x\) is the angle at the base of the triangle? Wait, maybe the triangle is equilateral? No, the sides are marked equal, but maybe the triangle has angles \(c = 36^{\circ}\) and \(d = 108^{\circ}\). Wait, let's think of the pentagon as a regular pentagon. In a regular pentagon, the triangle formed by two sides and a diagonal is isosceles with base angles \(36^{\circ}\) and vertex angle \(108^{\circ}\). Wait, no, the diagonal of a regular pentagon forms a triangle with two sides, and the base angles of that triangle are \(36^{\circ}\)? Wait, no, the interior angle of a regular pentagon is \(108^{\circ}\), and the triangle formed by two sides and a diagonal: the angle at the vertex of the pentagon is \(108^{\circ}\), and the two base angles of the triangle (formed by the diagonal) are \(\frac{180^{\circ}- 108^{\circ}}{2}=36^{\circ}\). Ah! So in the left figure, the angle \(c\) is \(36^{\circ}\) (the base angle of the isosceles triangle formed by the diagonal of the pentagon), and the angle \(d\) is \(108^{\circ}\) (the interior angle of the pentagon, which is the base angle of the isosceles trapezoid).

Step 2: Analyze the Right Figure (Regular Hexagon)

A regular hexagon has all sides equal and all interior angles equal to \(120^{\circ}\). The right figure is a regular hexagon with some diagonals. The triangle formed inside: in a regular hexagon, the diagonals that connect opposite vertices divide the hexagon into equilateral triangles? Wait, no, the side length of a regular hexagon is equal to the radius of the circumscribed circle. The angle \(g\): let's consider the triangle formed. The regular hexagon can be divided into six equilateral triangles with side length equal to the side of the hexagon. The angle \(h\): it's a supplementary angle to the interior angle of the hexagon. The interior angle of a regular hexagon is \(\frac{(6 - 2)\times180^{\circ}}{6}=120^{\circ}\), so the angle \(h\) (which is adjacent to the interior angle of the hexagon and forms a linear pair) is \(180^{\circ}- 120^{\circ}=60^{\circ}\). For the angle \(g\): looking at the triangle formed by the diagonals, since the hexagon is regular, the triangle is equilateral? Wait, no, the diagonals in the hexagon: if we connect the opposite vertices, we get a triangle with angles. Wait, the angle \(g\): in the regular hexagon, the triangle formed by two sides and a diagonal (or the intersection of diagonals) - actually, the angle \(g\) is \(60^{\circ}\)? Wait, no, let's think again. The regular hexagon has interior angle \(120^{\circ}\). The angle \(h\) is adjacent to the interior angle, so \(h = 180 - 120=60^{\circ}\). For the angle \(g\): the triangle formed by the diagonals in the hexagon is equilateral, so the angle \(g = 60^{\circ}\)? Wait, no, the intersection of the diagonals: in a regular hexagon, the diagonals that are not the long diagonals (connecting opposite vertices) form triangles. Wait, the right figure: the hexagon is regular, so all sides are equal, all interior angles are \(120^{\circ}\). The angle \(h\) is formed by a side of the hexagon and a line extending from the adjacent side, so it's a supplementary angle to the interior angle of the hexagon: \(h=180^{\circ}- 120^{\circ}=60^{\circ}\). The angle \(g\): looking at the triangle inside, since the hexagon is regular, the triangle is equilateral, so \(g = 60^{\circ}\)? Wait, no, the interior angle of the hexagon is \(120^{\circ}\), and the triangle formed by the diagonals: the angle \(g\) is equal to the angle of the equilateral triangle, which is \(60^{\circ}\). Wait, maybe the angle \(g = 60^{\circ}\) and \(h = 60^{\circ}\)? Wait, no, let's do it step by step.

For the right figure:

• A regular hexagon has each interior angle \(\theta=\frac{(6 - 2)\times180^{\circ}}{6}=120^{\circ}\).
• The angle \(h\) is adjacent to the interior angle of the hexagon, so they form a linear pair. Thus, \(h = 180^{\circ}-\ 120^{\circ}=60^{\circ}\).
• For the angle \(g\): the triangle formed by the diagonals in the hexagon is equilateral (since all sides of the hexagon are equal, and the diagonals in a regular hexagon create equilateral triangles). So the angle \(g = 60^{\circ}\).

Final Answers

• For the left figure:
• \(c=\boldsymbol{36^{\circ}}\)
• \(d=\boldsymbol{108^{\circ}}\)
• For the right figure:
• \(g=\boldsymbol{60^{\circ}}\)
• \(h=\boldsymbol{60^{\circ}}\)