Math Problem Statement

Find the value of x and the measure of each labeled angle in two geometric shapes.

Solution

Let's work through both problems in the image step by step.

Problem 15:

The quadrilateral shown is a trapezoid, and we are asked to find the value of xx and the measures of all labeled angles.

Given:

  • One angle is 9090^\circ (right angle).
  • One angle is labeled as xx^\circ.
  • The other angle is labeled as (x50)(x - 50)^\circ.

In any quadrilateral, the sum of the interior angles is 360360^\circ.

Let's call the angles as follows:

  • The right angle: 9090^\circ.
  • The opposite angle: xx^\circ.
  • The remaining two angles: (x50)(x - 50)^\circ and another 9090^\circ.

So, we can set up the equation:

90+90+x+(x50)=36090^\circ + 90^\circ + x^\circ + (x - 50)^\circ = 360^\circ

Simplifying the equation:

180+x+x50=360180^\circ + x + x - 50 = 360^\circ 2x50=1802x - 50 = 180^\circ 2x=2302x = 230^\circ x=115x = 115^\circ

Thus, x=115x = 115^\circ.

Angle measures:

  • x=115x = 115^\circ
  • (x50)=11550=65(x - 50) = 115^\circ - 50^\circ = 65^\circ
  • The right angles are 9090^\circ each.

Problem 16:

This figure is an isosceles triangle, and we are asked to find xx and the labeled angles.

Given:

  • One angle is (3x10)(3x - 10)^\circ.
  • Another angle is (x+40)(x + 40)^\circ.
  • The third angle (the top vertex) appears unlabeled, but since the sum of angles in a triangle is 180180^\circ, we can use that information.

The sum of the interior angles of a triangle is always 180180^\circ. So, we can set up the equation:

(3x10)+(x+40)+(x+40)=180(3x - 10)^\circ + (x + 40)^\circ + (x + 40)^\circ = 180^\circ

Simplifying the equation:

3x10+x+40+x+40=1803x - 10 + x + 40 + x + 40 = 180^\circ 5x+70=1805x + 70 = 180^\circ 5x=1105x = 110^\circ x=22x = 22^\circ

Angle measures:

  • x=22x = 22^\circ
  • (x+40)=22+40=62(x + 40) = 22^\circ + 40^\circ = 62^\circ
  • (3x10)=3(22)10=6610=56(3x - 10) = 3(22^\circ) - 10^\circ = 66^\circ - 10^\circ = 56^\circ

Summary:

  1. For problem 15: x=115x = 115^\circ, angles are 115115^\circ, 6565^\circ, and two 9090^\circ angles.
  2. For problem 16: x=22x = 22^\circ, angles are 6262^\circ, 6262^\circ, and 5656^\circ.

Would you like any further clarifications on these solutions?

Further Related Questions:

  1. How do you find the sum of the interior angles of any polygon?
  2. What are the properties of an isosceles triangle that can help solve angle problems?
  3. How can we verify angle relationships in a quadrilateral like a trapezoid?
  4. How do angle sum properties change for concave vs. convex polygons?
  5. What are the steps to solve problems involving supplementary and complementary angles?

Tip:

Remember that the sum of interior angles of any polygon can be found using the formula (n2)×180(n-2) \times 180^\circ, where nn is the number of sides.

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Trapezoid Properties
Isosceles Triangle Properties
Angle Sum of Polygons

Formulas

Sum of angles in a quadrilateral: 360°
Sum of angles in a triangle: 180°

Theorems

Angle Sum Theorem
Properties of Isosceles Triangles
Trapezoid Angle Properties

Suitable Grade Level

Grades 8-10