Question Number 204541 by necx122 last updated on 21/Feb/24
$${In}\:{a}\:{regular}\:{pentagon}\:{PQRST}\:,\:{PR} \\ $$$${intersects}\:{QS}\:{at}\:{O}.\:{Calculate}\:{ROS}? \\ $$
Commented by som(math1967) last updated on 21/Feb/24
$$\measuredangle{ROS}=\mathrm{72}° \\ $$
Commented by necx122 last updated on 21/Feb/24
$${please}\:{explain}\:{with}\:{diagram}\:{sir}. \\ $$$${Thank}\:{you}\:{immensely}. \\ $$
Commented by som(math1967) last updated on 21/Feb/24
Commented by som(math1967) last updated on 21/Feb/24
$${PQ}={QR}\:\therefore\angle{PRQ}=\frac{\mathrm{180}−\mathrm{108}}{\mathrm{2}}=\mathrm{36} \\ $$$$\angle{ORS}=\mathrm{108}−\mathrm{36}=\mathrm{72}° \\ $$$${QR}={RS}\:\angle{OSR}=\mathrm{36}° \\ $$$$\angle{ROS}=\mathrm{180}−\mathrm{72}−\mathrm{36}=\mathrm{72}° \\ $$$$ \\ $$