Given ΔUVW, segment UW is extended to point X
And the following angles are known
∠VWX=(7x-14)º
∠WUV=(2x+11)º
∠UVW=(3x-7)º
You have to find ∠VWX
First, let's make a sketch of the triangle and place the angles
To determine ∠VWX, first, we need to find the value of x.
For this, we have to apply the exterior angle theorem, which states that the measure of one exterior angle of a triangle is equal to the sum of the opposite interior angles, so that:
[tex]\begin{gathered} \angle\text{VWX}=\angle\text{WUV}+\angle\text{UVW} \\ (7x-14)º=(2x+11)º+(3x-7)º \end{gathered}[/tex]
From this expression, we can determine the value of x.
1) Erase the parentheses and on the right side of the equation order the like terms and simplify them:
[tex]\begin{gathered} 7x-14=2x+11+3x-7 \\ 7x-14=2x+3x+11-7 \\ 7x-14=5x+4 \end{gathered}[/tex]
2) Now you have to pass 5x to the left side of the expression and -14 to the right side. For this, apply the opposite operation to both sides of it.
For "5x" you have to subtract it and for "-14" you have to add it as follows:
[tex]\begin{gathered} 7x-5x-14=5x-5x+4 \\ 2x-14=4 \\ 2x-14+14=4+14 \\ 2x=18 \end{gathered}[/tex]
3) Divide both sides by 2 to determine the value of x
[tex]\begin{gathered} \frac{2x}{2}=\frac{18}{2} \\ x=9 \end{gathered}[/tex]
Now we can calculate the measure of ∠VWX
[tex]\begin{gathered} \angle\text{VWX}=7x-14 \\ \angle\text{VWX}=7\cdot9-14 \\ \angle\text{VWX}=63-14 \\ \angle\text{VWX}=49º \end{gathered}[/tex]
