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Question Number 102918 by otchereabdullai@gmail.com last updated on 11/Jul/20

Commented by bramlex last updated on 12/Jul/20

Commented by otchereabdullai@gmail.com last updated on 14/Jul/20

$$\mathrm{thanks}\mathrm{for}\mathrm{ur}\mathrm{time}\mathrm{sir}!$$

Answered by som(math1967) last updated on 11/Jul/20

$$\mathrm{\angle }\mathrm{QTS}=\mathrm{\angle }\mathrm{SPQ}=52°★$$$$\therefore \mathrm{\angle }\mathrm{QPS}=180-52=128°◼$$$$\mathrm{\angle }\mathrm{RQS}=\mathrm{\angle }\mathrm{RSQ}=\frac{180-128}{2}=26°◼◼$$$$\mathrm{\angle }\mathrm{RSQ}=\mathrm{\angle }\mathrm{SQT}=26°\mathrm{ans}$$$$\mathrm{\angle }\mathrm{SQP}=90°[\mathrm{semicirular}$$$$\mathrm{angle}]$$$$\therefore \mathrm{\angle }\mathrm{PQT}=\mathrm{\angle }\mathrm{SQP}-\mathrm{\angle }\mathrm{SQT}$$$$=90-26=64°\mathrm{ans}$$$$★\mathrm{subtain}\mathrm{on}\mathrm{same}\mathrm{segment}$$$$◼\mathrm{PQRS}\mathrm{cyclic}$$$$◼◼\mathrm{QR}=\mathrm{RS}$$$$$$

Commented by otchereabdullai@gmail.com last updated on 11/Jul/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}!\mathrm{Am}\mathrm{much}\mathrm{greatful}$$$$\mathrm{for}\mathrm{your}\mathrm{time}$$

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