Question

21. If the sides of a triangle are
produced in order, show that the
sum of the exterior angles so
formed is equal to four right
angles.​

Answer 1

$\large{ \bold{ \underline{ \underline{ \star{ \red{Question....}}}}}}$

$\large{ \mathrm{ \bullet{if \: the \: sides \: of \: a \: triangle \: are \: produced \: in \: order \: show \: that}}} \\ \large{ \mathrm{the \: sum \: of \: the \: exterior \: angles \: so \: formed \: is \: }} \\ \large{ \mathrm{equal \: to \: four \: right \: angles...}}$

$\large{ \bold{ \star{ \underline{ \underline{ \green{Solution...}}}}}}$

$\large{ \mathrm{ \star{ \underline{ \underline{ \pink { to \: prove}}}}}} \\ \\ \large{ \mathrm{ \dag{ \: sum \: of \: exterior \: angles = 360}}}$

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$\large{ \bold{ \star{ \underline{ \underline{ \red{Proof...}}}}}}$

$\large{ \mathrm{ \underline{ \purple{ \star{ \: \: refer \: to \: the \: attachement...}}}}}$

$\large{ \mathrm{as \: we \: know \: that...}} \\ \large{ \mathrm{ \boxed{ \red{sum \: of \: all \: angles \: of \: a \: triangle = 180}}}} \\ \large{ \mathrm{.....equation \: (1)}} \\ \\ : \large{ \mathrm{ \implies{ < \alpha + < \beta + < \gamma = 180}}} \\ \\ \large{ \mathrm{ \mapsto{exterior \: angle \: of \: < \alpha = 180 - \alpha }}} \\ \\ \large{ \mathrm{ \mapsto{ exterior \: angle \: of \: < \beta = 180 - \beta }}} \\ \\ \large{ \mathrm{ \mapsto{exterior \: angle \: of \: < \gamma = 180 - < \gamma }}} \\ \\ \large{ \mathrm{ \green{then \: sum \: of \: exterior \: angles \: of \: triangle}}} \\ \\ \large{ \mathrm{ = (180 - \alpha ) + (180 - \beta ) + (180 - \gamma )}} \\ \\ \large{ \mathrm{ = 180 - < \alpha + 180 - < \beta + 180 - < \gamma }} \\ \\ \large{ \mathrm{ \boxed{ \green{ = 180 + 180 + 180 - ( < \alpha + \beta + \gamma) }}}} \\ \large{ \mathrm{....equation \: (2)}} \\ \\ \sf{ \dag{ \underline{ \underline{ \red{putting \: equation \: (1) \: in \: (2)}}}}} \\ \\ \large{ \mathrm{ = 540 - 180}} \\ \\ \large{ \mathrm{ \boxed{ \purple{ = 360}}}} \\ \\ \large{ \therefore{ \pink{ \mathrm{hence \: proved...}}}}$

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