Question

please answer this question
if anyone know​

Answer 1

Answer:

first let's find BAQ

now we know that PAQ is equal to 180° and

PAB+BAQ=180°

58°+BAQ=180°. (PAB=58° given)

thus BAQ=122°

NOW LETS draw a perpendicular to the point O

from the point A

and now

PAB+BAO=90°(since A is perpendicular to O)

58°+BAO=90°

BAO=32°

NOW

BAO=ABO(Since OA=OB AND angle subtended by radius are equal )

thus ABO=32°=ABQ(BOTH ARE SAME ANGLE)

NOW

IN THE Triangle BAQ

BAQ+ABQ+AQB=180°

122°+32°+AQB=180°

thus AQB=26°

ABQ=32°andAQB=26°