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Question Number 97942 by pranesh last updated on 10/Jun/20

Answered by john santu last updated on 10/Jun/20

Answered by Aniruddha Ghosh last updated on 10/Jun/20

$$\mathbf{PQRS}\mathrm{is}\mathrm{a}\mathbf{circular}\mathbf{quadrilateral}.$$$$\mathrm{\angle }\mathbf{PQR}=180°-\mathrm{\angle }\mathbf{PSR}$$$$\therefore \mathrm{\angle }\mathbf{PSR}=180°-112°=68°$$$$\mathrm{now},\mathbf{outter}\mathrm{\angle }\mathbf{PTR}=2\times 68°=136°$$$$\mathrm{In}\mathbf{PQRT}\mathbf{quadrilateral},$$$$\mathrm{\angle }\mathbf{PQR}+\mathrm{\angle }\mathbf{QRT}+\mathrm{\angle }\mathbf{PTR}+\mathrm{\angle }\mathit{x}=360°$$$$112+28+136+\mathit{x}=360$$$$\mathit{x}=84°\left(\mathrm{this}\mathrm{is}\mathrm{your}\mathrm{answer}\right)$$$$$$

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