The problem asks to measure and record the angles formed at each intersection (points A, B, and C) of a triangle formed by three straight paths AB, BC, and AC. It also asks to measure and record the lengths of the straight paths (sides). Finally, it asks to capture the image of the triangle and paste or insert it in the space below. Since there is no triangle provided in the prompt, I will create a hypothetical triangle and measure its sides and angles.

1. Problem Description

The problem asks to measure and record the angles formed at each intersection (points A, B, and C) of a triangle formed by three straight paths AB, BC, and AC. It also asks to measure and record the lengths of the straight paths (sides). Finally, it asks to capture the image of the triangle and paste or insert it in the space below. Since there is no triangle provided in the prompt, I will create a hypothetical triangle and measure its sides and angles.

2. Solution Steps

First, let's consider a triangle with the following specifications:

* Side AB = 5 units

* Side BC = 6 units

* Side AC = 7 units

Using the Law of Cosines, we can find the angles of the triangle. The Law of Cosines states:

$c^2 = a^2 + b^2 - 2ab \cdot cos(C)$

where a, b, and c are the lengths of the sides of the triangle, and C is the angle opposite side c.

To find angle A (opposite side BC):

$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot cos(A)$

$6^2 = 5^2 + 7^2 - 2 \cdot 5 \cdot 7 \cdot cos(A)$

$36 = 25 + 49 - 70 \cdot cos(A)$

$36 = 74 - 70 \cdot cos(A)$

$70 \cdot cos(A) = 74 - 36$

$70 \cdot cos(A) = 38$

$cos(A) = \frac{38}{70}$

$A = arccos(\frac{38}{70})$

$A \approx 57.12^\circ$

To find angle B (opposite side AC):

$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot cos(B)$

$7^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cdot cos(B)$

$49 = 25 + 36 - 60 \cdot cos(B)$

$49 = 61 - 60 \cdot cos(B)$

$60 \cdot cos(B) = 61 - 49$

$60 \cdot cos(B) = 12$

$cos(B) = \frac{12}{60} = \frac{1}{5}$

$B = arccos(\frac{1}{5})$

$B \approx 78.46^\circ$

To find angle C (opposite side AB):

$AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot cos(C)$

$5^2 = 7^2 + 6^2 - 2 \cdot 7 \cdot 6 \cdot cos(C)$

$25 = 49 + 36 - 84 \cdot cos(C)$

$25 = 85 - 84 \cdot cos(C)$

$84 \cdot cos(C) = 85 - 25$

$84 \cdot cos(C) = 60$

$cos(C) = \frac{60}{84} = \frac{5}{7}$

$C = arccos(\frac{5}{7})$

$C \approx 44.42^\circ$

Note that the sum of angles in a triangle should be 180 degrees.

$A + B + C \approx 57.12 + 78.46 + 44.42 = 180^\circ$

3. Final Answer

Side AB = 5 units

Side BC = 6 units

Side AC = 7 units

Angle A $\approx 57.12^\circ$

Angle B $\approx 78.46^\circ$

Angle C $\approx 44.42^\circ$