Answer:
The measurement of ∠APB=80°
Step-by-step explanation:
Given:
In quadrilateral ABCD ∠C=100° and ∠D=60° .
Now, let ∠PAB=[tex]x[/tex] and ∠PBA=[tex]y[/tex]
As, sum of all interior angles of a quadrilateral is 360°.
∠A+∠B+∠C+∠D=360°
⇒[tex]2x+2y+100+60=360[/tex]
(as the line AP and BP bisect ∠A and ∠B respectively so, ∠A=2x and ∠B=2y)
⇒[tex]2x+2y+160=360[/tex]
subtracting 160 on both sides
⇒[tex]2x+2y=200[/tex]
dividing by 2 on both the sides
⇒[tex]\frac{2x}{2} +\frac{2y}{2}=\frac{200}{2}[/tex]
⇒[tex]x+y=100[/tex] ........(1)
Now, in triangle PAB:
Sum of interior angles of triangle is 180°.
∠PAB+∠PBA+∠APB=180°
Let ∠APB be z.
⇒[tex]x+y+z=180[/tex]
Putting the value from the above equation (1)
⇒[tex]100+z=180[/tex]
Subtracting 100 from both sides
⇒[tex]z=80[/tex]
⇒ ∠APB=80°
Therefore, the measurement of ∠APB=80° .
