Question

Prove that the angle in a segment shorter than a semicircle is greater than a right angle.

Answer 1

$$\begin{gathered} \stackrel{⏜}{QP}\mathrm{is}\mathrm{a}\mathrm{major}\mathrm{arc}\mathrm{and}\angle PSQ\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{formed}\mathrm{by}\mathrm{it}\mathrm{in}\mathrm{the}\mathrm{alternate}\mathrm{segment}. \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{an}\mathrm{arc}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{is}\mathrm{twice}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{by}\mathrm{it}\mathrm{at}\mathrm{any}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{alternate}\mathrm{segment}\mathrm{of}\mathrm{the}\mathrm{circle}. \\ \therefore 2\angle PSQ=m\left(\stackrel{⏜}{QP}\right) \\ \Rightarrow 2\angle PSQ=360°-m\left(\stackrel{⏜}{PQ}\right) \\ \Rightarrow 2\angle PSQ=360°-\angle POQ \\ \Rightarrow 2\angle PSQ=360°-180°\left(\because \angle POQ<180°\right) \\ \Rightarrow 2\angle PSQ>180° \\ \Rightarrow \angle PSQ>90° \end{gathered}$$

Thus, the angle in a segment shorter than a semi-circle is greater than a right angle.