PQ is a tangent to the circle at A. If angle PAB=58°, find angle ABQ

1.	PQ is a tangent to the circle at A. If angle PAB=58°, find angle ABQ and angle AQB
Answer» Join OA.\xa0clearly, OA {tex}\\perp{/tex}\xa0PAQ.{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}OAP = 90°{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}1 + 58° = 90°{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}1 = 90° - 58° = 32°In {tex}\\triangle{/tex}BOA,OA = OB. [radius of circle]\xa0Now,\xa0{tex}\\angle{/tex}PAB={tex}\\angle {/tex}ARB={tex}58^\\circ{/tex}[alternate\xa0angles are equal]{tex}\\angle{/tex}ABQ = 32° [as AO=OB,angles opposite to them must be equal]{tex}\\angle{/tex}PAB + {tex}\\angle{/tex}BAQ = 180°{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}BAQ = 180° - 58°= 122°In {tex}\\triangle{/tex}ABQ,{tex}\\angle{/tex}ABQ + {tex}\\angle{/tex}BAQ + {tex}\\angle{/tex}AQB = 180°\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\angle{/tex}{tex}AQB = 180° - 122° - 32° = 26°{/tex}

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