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 \[\Delta APB\] |
\[\angle APB\,=\,y\] |
\[\angle BAP\,=\,x\] |
\[\angle APB=\,180{}^\circ -(x+y)\] |
\[=180{}^\circ -100\] |
\[\angle APB=80{}^\circ \] |
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