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
$$\angle{PQR}=\frac{\mathrm{3}}{\mathrm{10}}×\mathrm{180}=\mathrm{54}°=\angle{PSR} \\ $$$${Now}\:\angle{QSR}=\frac{\mathrm{1}}{\mathrm{2}}\angle{PSR}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{54}°=\mathrm{27}° \\ $$
Commented by pieroo last updated on 27/Apr/18
$$\mathrm{thanks}\:\mathrm{boss} \\ $$