Question

Pls ans this question ​

Answer 1

hope you understand...

method is same but value is changed so you can put that value which you have instead of these, OK...

Answer 2

Answer:

$\huge{\boxed{\sf{\red{x = 270°}}}}$

Step-by-step explanation:

$\displaystyle{\sf{Draw\:a\:parallel\:line\:on\:M.}}$

$\displaystyle{\sf{ \angle PXA + \angle XMA = 180°}}$

|$\displaystyle{\sf{ \because the\:sum\:of\:co-interior\:angle\:is\:180°}}$

$\implies\displaystyle{\sf{ 50° + \angle XMA = 180°}}$

$\implies\displaystyle{\sf{ \angle XMA = 180°-50°}}$

$\implies\displaystyle{\sf{ \angle XMA = 130°}}$

$\displaystyle{\sf{ \angle AMY= \angle MYR = 120°}}$

|$\displaystyle{\sf{ \because Alternate\:Interior\:angle}}$

$\longrightarrow\displaystyle{\sf{ \angle AMY+ \angle XMA = (x-20)°}}$

$\longrightarrow\displaystyle{\sf{ 130°+ 120°= (x-20)°}}$

$\longrightarrow\displaystyle{\sf{ 250°= (x-20)^°}}$

$\longrightarrow\displaystyle{\sf{ 250°= (x-20)°}}$

$\longrightarrow\displaystyle{\sf{ 250°+ 20°= (x)°}}$

$\Rightarrow{\boxed{\sf{\red{ 270°= (x)°}}}}$

Hence, the value of x is equal to 270°

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