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Draw An angle of 135

Answer» Given: A ray OA.Required: To construct an angle of 1350\xa0at O.Steps of construction :\tProduce AO to A\'\xa0to form OA\'.\tTaking O as centre and some radius, draw an arc of a circle, which intersects OA at a point B and OA\'\xa0at a point B\'.\tTaking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.\tTaking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.\t\tDraw the ray OE passing through C.\tThen {tex}\\angle{/tex}EOA = 60o.\tDraw the ray OF passing through D.\tThen {tex}\\angle{/tex}FOE = 60o.\tTaking C and D as centres and with the radius more than {tex}1 \\over2{/tex}CD, draw arcs to intersect each other, say at G.\tDraw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE, i.e. {tex}\\angle{/tex}FOG = {tex}\\angle{/tex}EOG = {tex}1 \\over2{/tex}{tex}\\angle{/tex}FOE = {tex}1 \\over2{/tex}(60o) = 30o.\tThus {tex}\\angle{/tex}GOA = {tex}\\angle{/tex}GOE + {tex}\\angle{/tex}EOA = 30o + 60o = 90o\t{tex}\\angle{/tex}B\'OH = 90o\tTaking B\'\xa0and H as centres and with the radius more than {tex}1 \\over2{/tex}B\'H, draw arcs to intersect each other, say at I.\tDraw the ray OI. This ray OI is the bisector of the angle\xa0B\'OG i.e. {tex}\\angle{/tex}B\'OI = {tex}\\angle{/tex}GOI = {tex}1 \\over2{/tex}{tex}\\angle{/tex}B\'OG = {tex}1 \\over2{/tex}(90o) = 45o\t{tex}\\angle{/tex}IOA = {tex}\\angle{/tex}IOG + {tex}\\angle{/tex}GOA\t= 45o + 90o = 135o\tOn measuring the {tex}\\angle{/tex}IOA by protractor, find that {tex}\\angle{/tex}IOA = 135o.


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