Question

find angle P...........plzzzzzzzz

Answer 1

we know ,
sum of all angle of any quadrateral = 360°

so,
angle A + angle B + angle C + angle D = 360°

angle A + angle B + 105° + 55° = 360°

angle A + angle B = 360 - 160° = 200°

hence,
1/2( angle A + angle B) = 200°/2 = 100°

1/2( angle A + angle B ) = 100°

here ,
angle PAB = 1/2 angle A
angle PBA = 1/2 angle B

hence,

angle PAB + angle PBA = 1/2{ angle A + angle B } = 100° -------(1)

now,

PAB is a triangle so,

(angle PAB + angle PBA )+ angle P = 180°

100° + angle P = 180° { from eqn(1)

angle P = 180° - 100° = 80°

hence angle P = 80°

Answer 2

Given - Angle K = 55° and angle C = 105°

AP bisects angle A and BP bisects angle P.


As we know that the sum of all angles of any quadrilateral is 360°.

In the given figure provided A+B+C+K = 180°


we know the value of angle C and angle K , so let's put it.

A+B+55°+105° = 360°
or, A+B = 360° - 160°
or, A+B = 200°

So angle( A+B )= 200°

Angle PAB + Angle PBA = Angle (A+B)/2

[why? because AB bisects angle A and BP bisects angle B]


So angle PAB + angle PBA = 200°/2 = 100°

Therefore, Angle P = 180° - (angle PAB + Angle PBA)
[Why? angle sum property of triangle states that sum of all angles of triangles is 180°]


Angle P = 180°-100° = 80°

So, angle P is 180°

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