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Question Number 33880 by pieroo last updated on 26/Apr/18

Commented by pieroo last updated on 27/Apr/18

$$\mathrm{Please},\mathrm{i}\mathrm{really}\mathrm{need}\mathrm{help}\mathrm{with}\mathrm{the}\mathrm{above}$$

Answered by math1967 last updated on 27/Apr/18

$$\mathrm{\angle }PQR=\frac{3}{10}\times 180=54°=\mathrm{\angle }PSR$$$$Now\mathrm{\angle }QSR=\frac{1}{2}\mathrm{\angle }PSR=\frac{1}{2}\times 54°=27°$$

Commented by pieroo last updated on 27/Apr/18

$$\mathrm{thanks}\mathrm{boss}$$

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