Please anyone explain with steps the answer is 101°,79°

1.	Please anyone explain with steps the answer is 101°,79°
Answer» $\huge{ \mathbf{ \underline{ \underline{{ \color{goldenrod}{❁} \pink{Question:- }}}}}}$ If the two opposite angles formed at the POINT of ntersection of the DIAGONALS of a parallelogram are 2x+61° and 3x+ 41°, FIND the adjacent angle formed at the point. ✿Given:- the two opposite angles formed at the point of intersection of the diagonals of a parallelogram are 2x+61° and 3x+ 41°. ❀To find:- Adjacent angles formed at the point. ᯽︎Solution:- $\large{ \boxed{ \sf{ \underline{ \pink{Opposite \: angles \: of \: parallelogram \: are \: equal}}}}}$ $\purple{ \dag} \: \: \: \: \: \: \sf{So \: \angle \:AOD= \angle{COB}}$ $\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \red{ \implies} \bf \: 2x + 61 \degree = 3x + 41 \degree$ $\\ \: \: \: \: \: \: \: \: \: \: \pink{ : \implies} \bf \: 3x - 2x = 61 - 41 \degree$ $\\ \: \: \: \: \: \: \: \: \: \: \: \: \purple{: \implies} \bf \: x = 20 \degree$ Now by KEEPING the value of x we will find the measure of ∠AOD and ∠COB $\\ \: \: \: \: \: \: \: \: \: \: \: \orange{: ↦ } \angle \: \bf \: AOD = 2 \times 20 + 61 \degree \\ \: \: \: \: \: \color{aqua}{: ↦} \: 40 \degree + 61 \degree \\ \: \: \: \: \: \: \: \color{goldenrod}{ : ↦ } \: \: \: \: \: 101 \degree$ $\\ \: \: \: \: \: \: \: \red{ : ↦} \angle \bf \:COB \: = 3 \times 20 \degree + 41 ^{o} \\ \: \: \: \: \: \: \color{maroon} : ↦ \bf \: 60 ^{o} + 41 ^{o} \\ \: \: \: \: \: \: \: \: \color{aqua} : ↦ \: 101 ^{o}$ Adjacent angles are ∠AOD and ∠COB:- ↠∠BOC + ∠DOC=180° ↠3x+ 41° + ∠DOC=180° ↠101 +∠DOC=180° ↠∠DOC=180°-101 ↠∠DOC=79° $\sf \: ↠∠DOC=∠AOB \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: (opposite \: angles \: of \: a \: \|\| grm \: are \: equal)$ Similarly ∠AOB=79° Hence the answers ∠AOB=101° and ∠COD=79° $\color{yellow}━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━$

Answer»

$\huge{ \mathbf{ \underline{ \underline{{ \color{goldenrod}{❁} \pink{Question:- }}}}}}$

If the two opposite angles formed at the POINT of

ntersection of the DIAGONALS of a parallelogram

are 2x+61° and 3x+ 41°, FIND the adjacent angle

formed at the point.

✿Given:-

the two opposite angles formed at the point of intersection of the diagonals of a parallelogram are 2x+61° and 3x+ 41°.

❀To find:-

Adjacent angles formed at the point.

᯽︎Solution:-

$\large{ \boxed{ \sf{ \underline{ \pink{Opposite \: angles \: of \: parallelogram \: are \: equal}}}}}$

$\purple{ \dag} \: \: \: \: \: \: \sf{So \: \angle \:AOD= \angle{COB}}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \red{ \implies} \bf \: 2x + 61 \degree = 3x + 41 \degree$

$\\ \: \: \: \: \: \: \: \: \: \: \pink{ : \implies} \bf \: 3x - 2x = 61 - 41 \degree$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \purple{: \implies} \bf \: x = 20 \degree$

Now by KEEPING the value of x we will find the measure of ∠AOD and ∠COB

$\\ \: \: \: \: \: \: \: \: \: \: \: \orange{: ↦ } \angle \: \bf \: AOD = 2 \times 20 + 61 \degree \\ \: \: \: \: \: \color{aqua}{: ↦} \: 40 \degree + 61 \degree \\ \: \: \: \: \: \: \: \color{goldenrod}{ : ↦ } \: \: \: \: \: 101 \degree$

$\\ \: \: \: \: \: \: \: \red{ : ↦} \angle \bf \:COB \: = 3 \times 20 \degree + 41 ^{o} \\ \: \: \: \: \: \: \color{maroon} : ↦ \bf \: 60 ^{o} + 41 ^{o} \\ \: \: \: \: \: \: \: \: \color{aqua} : ↦ \: 101 ^{o}$

Adjacent angles are ∠AOD and ∠COB:-

↠∠BOC + ∠DOC=180°

↠3x+ 41° + ∠DOC=180°

↠101 +∠DOC=180°

↠∠DOC=180°-101

↠∠DOC=79°

$\sf \: ↠∠DOC=∠AOB \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: (opposite \: angles \: of \: a \: || grm \: are \: equal)$

Similarly ∠AOB=79°

Hence the answers ∠AOB=101° and ∠COD=79°

$\color{yellow}━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━$