Answer :
b) Angles x = 23.4 degrees, y = 97.4 degrees, and 2 = 59.3 degrees
To determine the angles of a triangle with sides X = 13 meters, Y = 15 meters, and Z = 6 meters, we can use the Law of Cosines. The Law of Cosines states that for a triangle with sides a, b, and c, and opposite angles A, B, and C, respectively:
[tex]$$c^2 = a^2 + b^2 - 2ab\cos(C)$$[/tex]
Step 1: Calculate angle opposite side Z (angle x)
[tex]$$\cos(x) = \frac{Y^2 + X^2 - Z^2}{2YX} = \frac{15^2 + 13^2 - 6^2}{2 \cdot 15 \cdot 13}$$[/tex]
[tex]$$\cos(x) = \frac{225 + 169 - 36}{390} = \frac{358}{390} = 0.9179$$[/tex]
Taking the inverse cosine to find the angle x:
[tex]$$x = \cos^{-1}(0.9179) \approx 23.4^\circ$$[/tex]
Step 2: Calculate angle opposite side X (angle y)
[tex]$$\cos(y) = \frac{Z^2 + Y^2 - X^2}{2ZY} = \frac{6^2 + 15^2 - 13^2}{2 \cdot 6 \cdot 15}$$[/tex]
[tex]$$\cos(y) = \frac{36 + 225 - 169}{180} = \frac{92}{180} = 0.5111$$[/tex]
Taking the inverse cosine to find the angle y:
[tex]$$y = \cos^{-1}(0.5111) \approx 59.3^\circ$$[/tex]
Step 3: Calculate angle opposite side Y (angle z)
The sum of the angles in a triangle is always 180 degrees.
[tex]$$z = 180^\circ - x - y$$[/tex]
[tex]$$z = 180^\circ - 23.4^\circ - 59.3^\circ \approx 97.4^\circ$$[/tex]
Therefore, the answer is (e). Cannot be determined.
We will use the law of cosines to find the angles. The formula is:
[tex]cos(C) = (a^2 + b^2 - c^2) / (2ab)[/tex]
Given sides X = 13, Y = 15, and Z = 6, we can compute:
- Angle x: cos(x) = [tex](13^2 + 15^2 - 6^2)[/tex] / (2 * 13 * 15) ≈ 0.049.
- Thus, x ≈ [tex]cos^{-1}[/tex](0.049) ≈ 87.2°.
- Angle y: cos(y) = [tex](15^2 + 6^2 - 13^2)[/tex] / (2 * 15 * 6) ≈ 0.981.
- Thus, y ≈ [tex]cos^{-1}[/tex](0.981) ≈ 11.3°
- Angle z: We know the sum of angles in a triangle is 180°, so z ≈ 180° - 87.2° - 11.3° ≈ 81.5°.
The angles do not match any of the given choices exactly. The most appropriate answer is: e) Cannot be determined.
