Question

27.point O is the common end point of the rays OA , OB , OC , OD , and OE. Show that <AOB + <BOC + <COD + <DOE + <EOA =360°
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Answer 1

Step-by-step explanation:

$\huge{ \red {\boxed{\bold {\green{Solution : }}}}}$

See image

$\large \underline \bold \blue{Have \: Given :- }$

$\bold{point \: O \: is \: the \: common \: end \: point } \\ \bold{ of \: the \: rays \: OA , OB , OC , OD , and OE}$

$\large \underline{ \bold{ \orange{To \: prove :- }}}$

$\angle{AOB} + \angle{BOC} + \angle{COD} + \angle{DOE} + \angle{EOA} =360°$

$\large \underline{ \bold{ \purple{Composition :-}}}$

Draw a ray OP opposite to ray OA.

$∠AOB , \angle{BOC} \: and \: \angle COP\: \\ are \: linear \: pair \: angles,$

$\therefore \: ∠AOB +∠BOC + ∠COP = 180° ----(1)$

$Also, ∠AOE, ∠DOE \: and \: ∠DOP \\ are \: linear \: pair \: angles,$

$\therefore \: ∠AOE + ∠DOE + ∠DOP= 180° —–––(2)$

By adding equations, (1) and (2), we get,

$∠AOB +∠BOC + ∠COP + AOE + ∠DOE + ∠DOP = 180° + 180°$

$\angle{AOB} + \angle{BOC} + ( \angle{COP} + \angle{DOP}) + \angle{AOE} + \angle{DOE } = 360°$

$∠AOB + ∠BOC + ∠COD + ∠DOE + ∠AOE = 360° </p><p>$

Hence Proved.

I hope it is helpful

Answer 2

Answer:

2 answer send kiya hai dear