question_answer

ABCD is a quadrilateral in which measures of \[\angle D\]and \[\angle C\] are \[60{}^\circ \], and \[100{}^\circ \] respectively. If the internal bisectors of \[\angle A\] and \[\angle B\] meet at P, then measure of \[\angle APB\] is

A) \[80{}^\circ \]               

B)        \[~90{}^\circ \]

C) \[100{}^\circ \]              

D)        \[110{}^\circ \]

Correct Answer: A

Solution :

Sol. [a]

Let, ZA = 2x, ZB =2y

As we know that sum of angles of a quadrilateral is \[360{}^\circ \]

\[\therefore \,\,\,\angle A\,+\angle B\,+\angle C\,+\angle D\,=360{}^\circ \]

\[2x+2y+60{}^\circ \,+100{}^\circ =360{}^\circ \]

\[2x+2y=200{}^\circ \]

\[x+y=100{}^\circ \]

In \[\Delta APB\]

\[\angle APB\,=\,y\]

\[\angle BAP\,=\,x\]

\[\angle APB=\,180{}^\circ -(x+y)\]

\[=180{}^\circ -100\]

\[\angle APB=80{}^\circ \]