Question

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Answer 1

$answer$

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.

Answer 2

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°