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Question Number 42331 by AR Akash last updated on 23/Aug/18

Answered by $@ty@m last updated on 23/Aug/18

$$\because PQ=QR=RS$$$$\mathrm{\&}OP=OQ=OR=OS$$$$\therefore △OPQ\cong △OQR\cong △ORSbySSS$$$$\therefore \mathrm{\angle }PQO=\mathrm{\angle }OQR={70}^o$$$$\therefore \mathrm{\angle }PAQ={20}^o.....Ans\left(i\right)$$$$Again$$$$\because OQ=OR$$$$\therefore \mathrm{\angle }ORQ={70}^o$$$$\Rightarrow \mathrm{\angle }QOR={40}^o$$$$\Rightarrow \mathrm{\angle }ROS={40}^o......Ans\left(iii\right)$$$$\mathrm{\angle }AOS={100}^o$$$$\mathrm{\angle }OAS={40}^o$$$$\therefore \mathrm{\angle }PAS={20}^o+{40}^o={60}^o.....Ans.\left(ii\right)$$$$$$

Commented by Vashu last updated on 24/Aug/18

$$how\mathrm{\angle }PAQcame?explain$$

Commented by $@ty@m last updated on 24/Aug/18

$$Angleofsemicircleisarightangle.$$$$\therefore \mathrm{\angle }APQ={90}^o$$$$\therefore \mathrm{\angle }PQA+\mathrm{\angle }PAQ={90}^o$$$$\Rightarrow {70}^o+\mathrm{\angle }PAQ={90}^o$$$$\Rightarrow \mathrm{\angle }PAQ={20}^o$$

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