Question

In the adjacent figure, the bisectors of ∠A and ∠B meet in a point P. D

If ∠C = 100° and ∠D = 60°, find the measure of ∠APB.

Answer 1

As we know that, Sum of all four angles of quadrilateral is 360^o .

\(\angle\)A + \(\angle\)B + \(\angle\)C + \(\angle\)D = 360°

\(\angle\)A + \(\angle\)B + 100° + 60° = 360°

\(\angle\)A + \(\angle\)B = 360° - 160°

= 200°

Now, according to question bisector of \(\angle\)A and \(\angle\)B meet in a point P and forms the triangle

PAB.

So,

1/2 \(\angle\)A + 1/2 \(\angle\)B = 200°/2

= 100°

As we know that, sum of all angles of triangle is 180°.

1/2 \(\angle\)A + 1/2 \(\angle\)B + \(\angle\)APB = 180°

100° + \(\angle\)APB = 180°

= 80°

So, \(\angle\)APB = 80°