Question Number 199339 by necx122 last updated on 01/Nov/23
Commented by necx122 last updated on 01/Nov/23
$${find}\:\angle{QSR}\:{where}\:{O}\:{is}\:{the}\:{centre} \\ $$$$\angle{OTQ}\:=\mathrm{15},\:\angle{TOR}=\mathrm{110} \\ $$
Commented by AST last updated on 01/Nov/23
$$\mathrm{20}° \\ $$
Commented by AST last updated on 01/Nov/23
$${POQ}=\mathrm{2}{PTQ}=\mathrm{30}°\Rightarrow{QOR}=\mathrm{40}°\Rightarrow{QSR}=\frac{{QOR}}{\mathrm{2}} \\ $$$$=\mathrm{20}° \\ $$
Answered by MM42 last updated on 01/Nov/23
$$\mathrm{20} \\ $$
Commented by necx122 last updated on 01/Nov/23
oh! I've seen my mistake, thank you, sirs.
angle at the centre is twice the angle at the circumference