Math Problem Statement

Find the perimeter of triangle XYZ. Round your answer to the nearest tenth if necessary. Figures are not necessarily drawn to scale.

Solution

The problem asks to find the perimeter of triangle XYZ\triangle XYZ, which is depicted on the right side of the image.

Information from the triangle:

  • Angle at Y=72Y = 72^\circ
  • Angle at Z=63Z = 63^\circ
  • Angle at X=45X = 45^\circ
  • Side XY=33XY = 33
  • Side YZ=25.5YZ = 25.5

To find the perimeter of XYZ\triangle XYZ, we need the length of side XZXZ. We can use the Law of Cosines to calculate XZXZ.

The Law of Cosines states:

XZ2=XY2+YZ22XYYZcos(Y)XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(\angle Y)

Substituting the known values:

  • XY=33XY = 33
  • YZ=25.5YZ = 25.5
  • Y=72\angle Y = 72^\circ

Now let's calculate XZXZ.The length of side XZXZ is approximately 34.934.9 units.

Now, to find the perimeter of XYZ\triangle XYZ, we add up the lengths of all three sides:

Perimeter=XY+YZ+XZ=33+25.5+34.9=93.4\text{Perimeter} = XY + YZ + XZ = 33 + 25.5 + 34.9 = 93.4

Thus, the perimeter of XYZ\triangle XYZ is approximately 93.4 units.

Would you like further details on any part of this process or help with related concepts?

Here are five related questions you might find useful:

  1. How is the Law of Cosines derived?
  2. How can the Law of Sines be applied to this triangle?
  3. Can the perimeter be found using a different method, such as the Law of Sines?
  4. What is the difference between the Law of Cosines and the Law of Sines?
  5. How would you calculate the area of XYZ\triangle XYZ?

Tip: Always ensure the angles you use are in degrees or radians, matching the mode of your calculator or function.

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Math Problem Analysis

Mathematical Concepts

Geometry
Triangles
Law of Cosines

Formulas

Law of Cosines: c^2 = a^2 + b^2 - 2ab * cos(C)

Theorems

Law of Cosines

Suitable Grade Level

Grades 10-12