Math Problem Statement

In the diagram below, ∠WXZ is a right angle. If m∠WXV is eleven more than three times m∠WXY and m∠YXV = 139°, find m∠YXZ.

Solution

Let's break down the problem:

We are given that:

  • WXZ\angle WXZ is a right angle, so mWXZ=90m\angle WXZ = 90^\circ.
  • mWXVm\angle WXV is 11 more than 3 times mWXYm\angle WXY, i.e., mWXV=3×mWXY+11m\angle WXV = 3 \times m\angle WXY + 11.
  • mYXV=139m\angle YXV = 139^\circ.

We need to find mYXZm\angle YXZ.

Let's proceed step by step.

  1. Let mWXY=xm\angle WXY = x.

  2. From the information given:

    mWXV=3x+11m\angle WXV = 3x + 11
  3. Since WXV\angle WXV, YXV\angle YXV, and WXZ\angle WXZ are angles at point XX that add up to a straight line (180°), we can write:

    mWXV+mYXV+mYXZ=180m\angle WXV + m\angle YXV + m\angle YXZ = 180^\circ

    Substituting the known values:

    (3x+11)+139+mYXZ=180(3x + 11) + 139^\circ + m\angle YXZ = 180^\circ

    Simplifying:

    3x+150+mYXZ=1803x + 150^\circ + m\angle YXZ = 180^\circ mYXZ=1803x150m\angle YXZ = 180^\circ - 3x - 150^\circ mYXZ=303xm\angle YXZ = 30^\circ - 3x
  4. Next, we use the fact that WXZ\angle WXZ is a right angle:

    mWXY+mYXZ=90m\angle WXY + m\angle YXZ = 90^\circ

    Substituting mYXZ=303xm\angle YXZ = 30^\circ - 3x:

    x+(303x)=90x + (30^\circ - 3x) = 90^\circ

    Simplifying:

    x+303x=90x + 30^\circ - 3x = 90^\circ 2x+30=90-2x + 30^\circ = 90^\circ 2x=60-2x = 60^\circ x=30x = -30^\circ

Since mWXY=xm\angle WXY = x, we find that mWXY=30m\angle WXY = -30^\circ. However, the value of xx is negative, which does not make sense in this context. There seems to be a contradiction in the setup; let me re-check it for potential clarifications.

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Math Problem Analysis

Mathematical Concepts

Geometry
Angle Relationships
Linear Pair
Algebra

Formulas

Sum of angles on a straight line: m∠1 + m∠2 + m∠3 = 180°
Right angle property: m∠WXZ = 90°

Theorems

Linear Pair Theorem
Right Angle Theorem

Suitable Grade Level

Grades 8-10