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1
Sum of angles in a quadrilateral =360°
So ∠A+∠B+∠C+∠D=360°
∠A+∠B+100°+60°=360°
∠A+∠B=200°
(∠A +∠B)/2=100°
PA bisects ∠A ,PB bisects ∠B
∠PAB=∠A/2 ,∠PBA=∠B/2
In Δ APB
∠PAB+∠PBA+∠APB=180°
∠A/2+∠B/2+∠APB=180°
∠APB=180°-100°=80°
So option D is the answer.
Answered by
8
Answer:
Option D , = 80°
Step-by-step explanation:
Sum of angles of a quadrilateral = 360°
Given :-
✓A and √B meets at point P
✓C = 100° and ✓D = 60°
To find :-
✓APB
Solution :-
✓A + ✓B + ✓C + ✓D = 360°
Now , filling the values we get,
✓A + ✓B + 100° + 60° = 360°
✓A + ✓B + 160° = 360°
✓A + ✓B = 200°
(✓A + ✓B)/2 = 100° ------(1)
PA bisects ✓A and PB bisects ✓B
✓PAB = ✓A/2
✓PBA = ✓B/2
Now , in ∆ APB,
By angle sum property,
✓PAB + ✓PBA + ✓APB = 180°
From equation 1 ,
✓A/2 + ✓B/2 + ✓APB = 180°
100° + ✓APB = 180°
✓APB = 80°
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