Answer:
The measure of the [tex]{\angle}VPL[/tex] is [tex]80^{\circ}[/tex]
Step-by-step explanation:
We know that the measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.
Now, Let
x is the measure of arc IV
and y is the measure of arc PK, then
[tex]m{\angle}ISV={\frac{1}{2}}(x+y)[/tex]
Substituting the given values, we get
[tex]135^{\circ}={\frac{1}{2}(140^{\circ}+y)[/tex]
[tex]270^{\circ}=140^{\circ}+y[/tex]
[tex]y=130^{\circ}[/tex]
Thus, The measure of arc PK is [tex]130^{\circ}[/tex].
Also, we know that the inscribed angle measures half that of the arc comprising
, thus
Let
z is the measure of arc VK
and y is the measure of arc PK, then
[tex]m{\angle}VPL={\frac{1}{2}(z+y)[/tex]
Substituting the values, we get
[tex]m{\angle}VPL={\frac{1}{2}}(30+130)=80^{\circ}[/tex]
Hence, the measure of the [tex]{\angle}VPL[/tex] is [tex]80^{\circ}[/tex].
