In triangle XYZ, |XY|= 8 cm and |YZ|=6m.
The area of triangle XYZ is 12cm^2. The |<PQR|= 30 degrees, 150 degrees.
Find the two possible values of |XZ, correct 1 decimal place
To find the two possible values of |XZ in triangle XYZ, we can use the Law of Cosines. This law states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of the magnitudes of those two sides multiplied by the cosine of the included angle.
Let's denote |XZ as "a". Given that |XY = 8 cm, |YZ = 6 cm, and the area of triangle XYZ is 12 cm^2, we can now proceed as follows:
1. Calculate the magnitude of angle <XYZ using the area formula:
Area = 1/2 * |XY| * |YZ| * sin(<XYZ)
12 cm^2 = 1/2 * 8 cm * 6 cm * sin(<XYZ)
sin(<XYZ) = 1 cm^2 / (8 cm * 6 cm)
sin(<XYZ) ≈ 0.0833
2. With the magnitude of <XYZ determined, we can calculate the magnitude of <XZY as the supplement of <XYZ:
<XZY = 180° - <XYZ
<XZY = 180° - arcsin(0.0833) ≈ 172.21°
3. Now we can apply the Law of Cosines to find the values of |XZ:
a^2 = |XY|^2 + |YZ|^2 - 2 * |XY| * |YZ| * cos(<XYZ)
a^2 = 8 cm^2 + 6 cm^2 - 2 * 8 cm * 6 cm * cos(30°)
a^2 ≈ 64 cm^2 + 36 cm^2 - 96 cm^2 * 0.866
a^2 ≈ 27.62 cm^2
a ≈ √27.62 ≈ 5.26 cm
b^2 = |XY|^2 + |YZ|^2 - 2 * |XY| * |YZ| * cos(<XZY)
b^2 = 8 cm^2 + 6 cm^2 - 2 * 8 cm * 6 cm * cos(172.21°)
b^2 ≈ 64 cm^2 + 36 cm^2 - 96 cm^2 * (-0.982)
b^2 ≈ 191.55 cm^2
b ≈ √191.55 ≈ 13.85 cm
Therefore, the two possible values of |XZ in triangle XYZ are approximately 5.3 cm (correct to 1 decimal place) and 13.9 cm (correct to 1 decimal place).