SOLUTION
Consider the image given below.
From the image above,
[tex]\begin{gathered} \angle V=(x-2)^0 \\ \angle W=(x-2)^0 \\ \angle X=(4x+4)^0 \end{gathered}[/tex]To find the measures of each angle, we need to obtain the value of small x in the diagram.
Applying the rule: Sum of angle in a triangle is 180 degrees
[tex]\angle V+\angle W+\angle X=180^0[/tex]Then substitute the given expression, we have
[tex]\begin{gathered} (x-2)^0+(x-2)^0+(4x+4)^0=180^0 \\ Then \\ x-2+x-2+4x+4=180^0 \end{gathered}[/tex]Collect like terms and add
[tex]\begin{gathered} x+x+4x-2-2+4=180^0 \\ 6x-4+4=180^0 \\ \text{then} \\ 6x=180 \end{gathered}[/tex]Divide both sides by 6, we have
[tex]\begin{gathered} \frac{6x}{6}=\frac{180}{6} \\ hence \\ x=30 \end{gathered}[/tex]hence, x=30
Then substitute the value of x to obtain the measure of each angles.
[tex]\begin{gathered} \angle V=(x-2)^0 \\ \text{Then x=30} \\ \angle V=30-2=28^0 \end{gathered}[/tex]Since the triangle is an issoceles triangle, then
[tex]\angle W=28^0[/tex]Hence
[tex]\begin{gathered} \angle X=(4x+4)^0 \\ \angle X=(4\times30+4)^0=(120+4)=124^0 \\ \text{hence} \\ \angle X=124^0 \end{gathered}[/tex]Therefore
Answer : ∠V= 28°, ∠W=28°, ∠X=124°
