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°​

Answer 1

Step-by-step explanation:

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

See image

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

$\bold{Rays \: OA, OB, OC, OD and OE} \\ \bold{have \: the \: common \: endpoint \: O.}$

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

$∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°$

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

Draw a ray OP opposite to ray OA.

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

$∴∠AOB+∠BOC+∠COP=180°−−−−(1)$

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

$∴∠AOE+∠DOE+∠DOP=180° —–––(2) </p><p>$

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

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

$∠AOB+∠BOC+(∠COP+∠DOP)+∠AOE+∠DOE=360°$

$∠AOB + ∠BOC + ∠COD + ∠DOE + ∠AOE = 360°$

$\large{ \bold{ \pink{Hence \: Proved.}}}$

I hope it is helpful

Answer 2

Step-by-step explanation:

RaysOA,OB,OC,ODandOE

havethecommonendpointO.

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

Toprove:−

∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°

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

Composition:−

Draw a ray OP opposite to ray OA.

\begin{gathered} \bold{∠AOB , \angle{BOC} \: and \: \angle COP} \\ \bold{are \: linear \: pair \: angles,}\end{gathered}

∠AOB,∠BOCand∠COP

arelinearpairangles,

∴∠AOB+∠BOC+∠COP=180°−−−−(1)∴∠AOB+∠BOC+∠COP=180°−−−−(1)

\begin{gathered} \bold{Also, ∠AOE, ∠DOE \: and \: ∠DOP} \\ \bold{are \: linear \: pair \: angles,}\end{gathered}

Also,∠AOE,∠DOEand∠DOP

arelinearpairangles,

∴∠AOE+∠DOE+∠DOP=180° —–––(2) < /p > < p >∴∠AOE+∠DOE+∠DOP=180°—–––(2)</p><p>

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

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

∠AOB+∠BOC+(∠COP+∠DOP)+∠AOE+∠DOE=360°∠AOB+∠BOC+(∠COP+∠DOP)+∠AOE+∠DOE=360°

∠AOB + ∠BOC + ∠COD + ∠DOE + ∠AOE = 360°∠AOB+∠BOC+∠COD+∠DOE+∠AOE=360°

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