Question

In the adjoining figure, ABCD is a quadrilateral such that ∠D + ∠C = 100°. The bisectors of ∠A and ∠B meet at ∠P. Determine ∠APB

Answer 1

Answer:since angle a +b+c+d=360

Angle a+b =260

Ap and pb are bisectors ,so

The complete Angle a=130 and b=130

In triangle apb

A+p+b=180

A=65 and b=65

Therefore p= 50

Step-by-step explanation:

Answer 2

The value of $\angle APB$ is $50\ degree$.

Step-by-step explanation:

Given,

$ABCD$ quadrilateral such that $(\angle D+\angle C)=100\ degree$

For $ABCD$ quadrilateral,

$(\angle D+\angle C+\angle A+\angle B)=360\ degree$

⇒$(\angle A+\angle B)=(360-100)\ degree$

⇒$(\angle A+\angle B)=260\ degree$

From $\triangle ABP$,

$\frac{\angle A}{2} +\frac{\angle B}{2} +\angle P=180$

⇒$\frac{1}{2}(\angle A+\angle B) +\angle P=180$

⇒$\angle P=180-\frac{1}{2}(\angle A+\angle B)$

⇒$\angle P=180-\frac{1}{2}\times 260$

⇒$\angle P=(180-130)$

⇒$\angle P=50\ degree$

So, The value of $\angle APB$ is $50\ degree$.