Question Number 5431 by Rasheed Soomro last updated on 14/May/16
Commented by Yozzii last updated on 14/May/16
$${What}\:{does}\:\boldsymbol{\mathrm{d}}\:{measure}? \\ $$
Commented by Rasheed Soomro last updated on 14/May/16
$$\mathrm{d}\:\mathrm{is}\:\mathrm{a}\:\mathrm{name}\:\mathrm{of}\:\mathrm{semi}-\mathrm{circle}. \\ $$$$\mathrm{It}\:\mathrm{has}\:\mathrm{no}\:\mathrm{effect}\:\mathrm{on}\:\mathrm{problem}. \\ $$$$\mathrm{I}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{delete}\:\mathrm{it}\:\mathrm{but}\:\mathrm{couldn}'\mathrm{t}. \\ $$
Commented by Yozzii last updated on 14/May/16
$${No}\:{problem}.\: \\ $$
Commented by Yozzii last updated on 15/May/16
$${I}'{ll}\:{try}\:{it}\:{soon}.\:{I}'{ve}\:{been}\:{busy}. \\ $$
Commented by Rasheed Soomro last updated on 16/May/16
$$\underset{\boldsymbol{\mathrm{D}}} {\overset{\boldsymbol{\mathrm{G}}} {\mathcal{OO}}} \\ $$
Commented by Yozzii last updated on 16/May/16
$${Let}\:{the}\:{square}\:{be}\:{of}\:{side}\:{length}\:\mathrm{2}{r}. \\ $$$${Hence},\:{the}\:{radius}\:{of}\:{the}\:{semicircle}\:{is}\:{r}. \\ $$$${Since}\:{the}\:{quadrilateral}\:{in}\:{the}\:{figure} \\ $$$${is}\:{a}\:{square},\:{its}\:{diagonals}\:{are}\:{mutually} \\ $$$${orthogonal}.\:{So},\:{if}\:{O}\:{is}\:{the}\:{centre} \\ $$$${of}\:{the}\:{semicircle},\:{X}\:{is}\:{the}\:{vertex}\:{of} \\ $$$${the}\:{square}\:{where}\:{angle}\:{x}\:{arises},\:{and} \\ $$$${J}\:{is}\:{one}\:{of}\:{the}\:{end}\:{points}\:{of}\:{the}\:{arc} \\ $$$${of}\:{the}\:{semicircle},\:{then}\:\bigtriangleup{OJX}\:{is} \\ $$$${right}−{angled}.\:{By}\:{symmetry}\:{of}\:{the} \\ $$$${figure}\:\angle{OXJ}=\frac{{x}}{\mathrm{2}}\:{and}\:\angle{JOX}=\mathrm{90}°.\:{Sincr}\:{OJ}={r}\:{and} \\ $$$${OX}={r}\sqrt{\mathrm{2}}\:\Rightarrow{cos}\frac{{x}}{\mathrm{2}}=\frac{{r}\sqrt{\mathrm{2}}}{\sqrt{{r}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\mathrm{2}{cos}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\approx\mathrm{70}.\mathrm{53}° \\ $$$$ \\ $$$$ \\ $$$$ \\ $$