Question

Help me get this question correct

Answer 1

Basic theorems in geometry:

`\mathrm{1.\ Sum\ of\ interior\ angles\ of\ a\ triangle\ is\ 180^o.}\\\mathrm{2.\ Sum\ of\ all\ exterior\ angles\ of\ a\ triangle\ is\ 360^o.}\\\mathrm{3.\ An\ exterior\ angle\ of\ a\ triangle\ is\ equal\ to\ the\ sum\ of\ the\ opposite\ interior}\\\mathrm{\ \ \ angles.}\\\mathrm{4.\ Corresponding\ angles\ are\ equal.}\\\mathrm{5.\ Alternate\ angles\ are\ equal.}\\\mathrm{6.\ Vertically\ opposite\ angles\ are\ equal.}\\\mathrm{7.\ Cointerior\ angles\ are\ supplementary.}`

`\mathrm{8.\ Sum\ of\ angles\ in\ straight\ line\ is\ 180^o.}`

Answer:

`\bold{Method\ 1:}`

`\angle2= \angle6=60^o\ \ \ \ [\mathrm{Corresponding\ angles\ are\ equal.}]\\\mathrm{\angle1+\angle2=180^o}\ \ \ \ [\mathrm{Sum\ of\ angles\ in\ straight\ line\ is\ 180^o.}]\\\mathrm{or,\ \angle1+60^o=180^o}\\\mathrm{or,\ \angle1=120^o}`

`\bold{Method\ 2:}\\(i)\ \ \mathrm{\angle6+\angle3=180^o\ \ \ [\mathrm{Sum\ of\ co-interior\ angles\ is\ 180^o.}]}\\\mathrm{or,\ \ 60^o+\angle3=180^o}\\\mathrm{or,\ \ \angle3=120^o}\\(ii)\ \ \mathrm{\angle3=\angle1=120^o\ \ \ \ [\mathrm{Vertically\ opposite\ angles\ are\ equal.}]}`

`\bold{Method\ 3:}\\(i)\ \ \mathrm{\angle6=\angle4=60^o\ \ \ [Alternate\ angles\ are\ equal.]}\\(ii)\ \ \mathrm{\angle1+\angle4=180^o\ \ \ [\mathrm{Sum\ of\ angles\ in\ straight\ line\ is\ 180^o.}]}\\\mathrm{or,\ \ \angle1+60^o=180^o}\\\mathrm{or,\ \ \angle1=120^o}`