Question

in the adjacent figure the bisectors of angle A and Angle B meet in a point P if angle c = 100 degree and Angle d = 60 degree find the measure of angle APB



( please give right answer ) ​

Answer 1

Answer:

60

∘

+100

∘

+∠A+∠B=360

∘

⇒∠A+∠B=200

∘

⇒

2

∠A+∠B

=100

∘

⇒∠BAP+∠ABP=100

∘

But ∠BAP+∠ABP+∠APB=180

∘

⇒∠APB=80

∘

Answer 2

Answer:-

∠APB = 80°

Given:-

• ∠C = 100°

• ∠D = 60°

• ∠A and ∠B bisectors meet at point P

To Find:-

• ∠APB

Solution:-

Here, ABCD is a quadrilateral and we know that sum of all angles of quadrilateral is 360°.

So, ∠A + ∠B + ∠C + ∠D = 360°

 ⇒ ∠A + ∠B + 100° + 60° = 360°

 ⇒ ∠A + ∠B = 360° - 160°

 ⇒ ∠A + ∠B = 200°                  -------- (i)

Since, It is given that ∠A and ∠B is bisected.

So, $\dfrac{1}{2}\angle A$ = ∠PAB    and,

$\dfrac{1}{2}\angle B$ = ∠PBA  

From above we get,

$\dfrac{1}{2}\bigg(\angle A + \angle B \bigg)$  = ∠PAB + ∠PBA                    [ From equation (i) ]

$\dfrac{1}{2} \times 200$   ⇒ 100°  = ∠PAB + ∠PBA

Therefore, ∠PAB + ∠PBA  = 100°                 ----- (ii)

So, In ΔAPB

∠PAB + ∠PBA + ∠APB = 180°             [ Angle sum property of triangle ]

( ∠PAB + ∠PBA  ) + ∠APB = 180°

100°  +  ∠APB = 180°                           [ From equation (ii) ]

∠APB = 180° - 100°

∠APB = 80°

Hence, ∠APB = 80°

Some Important Terms:-

• Sum of all angles of quadrilateral is 360°

• Sum of all angles of triangle is 180°

• An Exterior angle of a triangle is equal to sum of its opposite interior angles.