Question

please help me Q.27 please ​

Answer 1

Step-by-step explanation:

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

See image

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

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

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

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

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

Draw a ray OP opposite to ray OA.

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

$\angle{AOB}+ \angle{BOC}+ \angle{COP }= 180° - - - -(1)$

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

$\angle{AOE}+ \angle{DOE}+ \angle{DOP}=180° - - - - (2)$

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

$\angle{AOB}+ \angle{BOC}+\angle{COP}+\angle{ AOE} \angle{DOE}+\angle{DOP}= 180°+180°$

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

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

Hence proved

I hope it is helpful

Answer 2

Step-by-step explanation:

RaysOA,OB,OC,ODandOE

havethecommonendpointO.

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

Toprove:−

\angle{AOB}+\angle{BOC}+\angle{COD} + \angle{DOE}+\angle{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{\angle{AOB}, \angle{BOC} \: and \angle{COP}} \\ \bold{ are \: linear \: pair \: angles,}\end{gathered}

∠AOB,∠BOCand∠COP

arelinearpairangles,

\angle{AOB}+ \angle{BOC}+ \angle{COP }= 180° - - - -(1)∠AOB+∠BOC+∠COP=180°−−−−(1)

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

Also,∠AOE,∠DOEand∠DOP

arelinearpairangles,

\angle{AOE}+ \angle{DOE}+ \angle{DOP}=180° - - - - (2)∠AOE+∠DOE+∠DOP=180°−−−−(2)

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

\angle{AOB}+ \angle{BOC}+\angle{COP}+\angle{ AOE} \angle{DOE}+\angle{DOP}= 180°+180°∠AOB+∠BOC+∠COP+∠AOE∠DOE+∠DOP=180°+180°

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

\angle{AOB}+ \angle{BOC}+ \angle{COD}+\angle{DOE}+\angle{AOE} = 360°∠AOB+∠BOC+∠COD+∠DOE+∠AOE=360°