Question

answer plizzz .......​

Answer 1

$\huge\beta\orange{Answer}$

Y=30° bcz We know that The angles on linear pair are always = 180°

SO,

90° + 3y = 180°

3y = 180°- 90°

3y = 90°

y= 90° ÷ 3

Y = 30°

Hope it helps...

Answer 2

$\Large{\underbrace{\sf{\pink{Required\:Solution:}}}}$

Given thαt,

• $\displaystyle{\sf{ \angle AOC = \angle DOB = 3y}}$ reαson : verticαlly opposite αngle.
• $\displaystyle{\sf{ \angle FOD = \angle BOE = 90^{\circ}}}$ reαson : verticαlly opposite αngle.
• $\displaystyle{\sf{ \angle COE= \angle FOD= 30^{\circ}}}$ reαson : verticαlly opposite αngle.

◾️We need to find the vαlue of y.

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According to the question,

$\longrightarrow{\sf{ \angle AOF +\angle FOD+ \angle DOB=180^{\circ}}}$

• Reason - Linear pair axiom.
• Substitute the values.

$\longrightarrow{\sf{ 90^{\circ}+30^{\circ}+3y=180^{\circ}}}$

$\longrightarrow{\sf{ 120^{\circ}+3y=180^{\circ}}}$

• Transpose 120° to R.H.S.

$\longrightarrow{\sf{ 3y=180^{\circ}-120^{\circ}}}$

$\longrightarrow{\sf{ 3y=60^{\circ}}}$

$\longrightarrow{\sf{ y= \dfrac{60^{\circ}}{3}}}$

$\longrightarrow{\sf{ y= \dfrac{\cancel{60^{\circ}}}{\cancel{3}}}}$

$\large{:}\implies{\boxed{\sf{\pink{ y = 20^{\circ}}}}}$

Hence, the value of y is 20°

And we are done! :D

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