Question Number 6806 by Tawakalitu. last updated on 27/Jul/16
Answered by Yozzii last updated on 28/Jul/16
$${We}\:{have}\:{that}\:\angle{QSR}=\mathrm{40}°.\:{Let}\:\omega\:{be}\: \\ $$$${circle}\:{PQRS}.\:{Points}\:{S}\:{and}\:{P}\:{lie}\:{on} \\ $$$$\omega\:{on}\:{the}\:{same}\:{side}\:{of}\:{line}\:{QR}.\: \\ $$$$\Rightarrow\angle{QSR}=\angle{QPR}\:\therefore\:\angle{QPR}=\mathrm{40}°. \\ $$$${TU}\:{is}\:{a}\:{straight}\:{line},\:{so}\: \\ $$$$\angle{TPS}+\angle{SPR}+\angle{QPR}=\mathrm{180}° \\ $$$$\Rightarrow\angle{SPR}=\mathrm{180}°−\mathrm{40}°−\mathrm{74}°=\mathrm{66}° \\ $$$${Angles}\:\angle{SPR}\:{and}\:\angle{SQR}\:{are}\:{in}\:{the} \\ $$$${same}\:{segment}\:{of}\:\omega\Rightarrow\:\angle{SQR}=\angle{SPR}=\mathrm{66}°. \\ $$$${We}\:{know}\:{that}\:\angle{RQU}=\mathrm{68}° \\ $$$${Now},\:\angle{PQS}+\angle{SQR}+\angle{RQU}=\mathrm{180}° \\ $$$$\Rightarrow\angle{PQS}=\mathrm{180}°−\mathrm{66}°−\mathrm{68}°=\mathrm{46}°. \\ $$$${Since}\:\angle{PQS}\:{and}\:\angle{PRS}\:{lie}\:{in}\:{the}\:{major} \\ $$$${segement}\:{PQRS},\:\angle{PRS}=\angle{PQS}=\mathrm{46}°. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 28/Jul/16
$${Wow}\:{thanks}. \\ $$