Question

In the given figure , AB is a side of a regular 6-sided polygon and AC is a side of a regular 8-sided polygon inscribed in a circle with the centre O. Then (i) $\angle AOB$ (ii) $\angle ACB$ (iii) $\angle ABC$ are respectively

• A. ${70}^o,{40}^o\&{44.5}^o.$
• B. ${60}^o,{30}^o\&{22.5}^o.$
• C. ${20}^o,{60}^o\&{11.5}^o.$
• D. ${80}^o,{50}^o\&{28.5}^o.$

Answer 1

The correct option is B ${60}^o,{30}^o\&{22.5}^o.$

$Given-ABisasideofaregularhexagonandACisasideofaregularoctagon.OisthecentreofacircleofwhichAB\&ACarechords.OA,OB\&BChavebeenjoined.Tofindout-\left(i\right)\angle AOB=?\angle ACB=?\angle ABC=?Solution-WejoinOC.ABisasideofaregularhexagonanditisalsoachordofthecircle.i.ethehexagonhasbeeninscribedinthegivencircle.\therefore Oisthecentreofthehexagonaswellasofthecircle.Weknowthattheanglesubtendedbyasideofaregularpolygonis\frac{{360}^o}{no.ofsides}..........\left(i\right)Hereno.ofsides=6.\therefore Anglesubtendedbythesideofofthehexagonis\angle AOB=\frac{{360}^o}{6}={60}^o......\left(i\right)SimilarlyACisasideofaregularoctagonanditisalsoachordofthecircle.i.etheoctagonhasbeeninscribedinthegivencircle.\therefore Oisthecentreoftheoctagonaswellasofthecircle.\therefore by\left(i\right),\angle AOC=\frac{{360}^o}{8}={45}^o(hereno.ofsides=8).NowACsubtends\angle ABCtothecircumferenceand\angle AOCtothecentre.Weknowthattheangle,subtendedbyachordofacircleatthecentre,isdoubletheanglesubtendedbythesamechordatthecircumferenceofthecircle..........\left(a\right)\therefore \angle ABC=\frac{1}{2}\angle AOC=\frac{1}{2}\times {45}^o={22.5}^o.........\left(iii\right)SimilarlyABsubtends\angle ACBtothecircumferenceand\angle AOBtothecentre.\therefore By\left(a\right)\angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {60}^o={30}^o.........\left(ii\right)So\angle AOB,\angle ACB\&\angle ABCarerespectively{60}^o,{30}^o\&{22.5}^o.Ans-OptionB.$