Question

find the range of values for x.
14
62°
(2x + 10)°
15
? < x <

Explanation:

Step1: Recall triangle - inequality theorem

In a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Also, we can use the angle - side relationship. In a triangle, the larger angle is opposite the longer side.
Let's consider the two triangles formed by the diagonal of the quadrilateral.
First, consider the angle - side relationship. We know that in a triangle, if we assume the two triangles formed by the diagonal.
We have two sides of lengths 14 and 15.
The angle opposite the side of length 15 is \(62^{\circ}\) and the angle opposite the side of length 14 is \((2x + 10)^{\circ}\)
Since the side of length 15> the side of length 14, then \(62^{\circ}>(2x + 10)^{\circ}\) (by the angle - side relationship in a triangle: larger side is opposite the larger angle)

Step2: Solve the inequality \(62>2x + 10\)

Subtract 10 from both sides of the inequality: \(62-10>2x+10 - 10\), which gives \(52>2x\)
Divide both sides by 2: \(\frac{52}{2}>x\), so \(x < 26\)
Also, since an angle measure must be non - negative, \(2x+10>0\)
Subtract 10 from both sides: \(2x>- 10\)
Divide both sides by 2: \(x>-5\)

Answer:

\(-5