Question

Find the value help plz ​

Answer 1

To Find :-

• The value of x,y,z & w

Solution :-

We are given a quadrilateral whose 3 angles are,

• 60°
• 80°
• 120°

We know,

• In a quadrilateral angles add up to 360°

• In a linear pair the Angles add up to 180°

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

On the way to finding w:-

$:\implies \displaystyle\sf 60^{\circ}+80^{\circ}+120^{\circ}+w' = 360^{\circ}$

$:\implies \displaystyle\sf 260 + w' = 360^{\circ}$

$:\implies \displaystyle\sf w' = 360-260$

$:\implies \displaystyle\sf w' = 100^{\circ}$

So then w will be,

$:\implies \displaystyle\sf w+w' = 180^{\circ} \:\:\{ Linear \ Pair\}$

$:\implies \displaystyle\sf w+100 = 180^{\circ}$

$:\implies \displaystyle\sf w = 180-100$

$\implies{\underline{\boxed{\frak{\red{w= 80^{\circ}}}}}}\;\bigstar$

Similarly x will be,

$:\implies \displaystyle\sf x'+x = 180^{\circ}\:\:\{ Linear \ Pair\}$

$:\implies \displaystyle\sf 120^{\circ}+x = 180^{\circ}$

$:\implies \displaystyle\sf x = 180-120$

$\implies{\underline{\boxed{\frak{\red{x= 60^{\circ}}}}}}\;\bigstar$

Value of y will be,

$:\implies \displaystyle\sf y'+y = 180^{\circ} \:\:\{ Linear \ Pair\}$

$:\implies \displaystyle\sf 80^{\circ}+y = 180^{\circ}$

$:\implies \displaystyle\sf y = 180-80$

$\implies{\underline{\boxed{\frak{\red{y= 100^{\circ}}}}}}\;\bigstar$

And finally the value of z will be,

$:\implies \displaystyle\sf z'+z = 180^{\circ}\:\:\{ Linear \ Pair\}$

$:\implies \displaystyle\sf 60^{\circ}+z = 180^{\circ}$

$:\implies \displaystyle\sf z = 180-60$

$\implies{\underline{\boxed{\frak{\red{z= 120^{\circ}}}}}}\;\bigstar$

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Answer 2

$\huge\color{red}{To \: Find :-}$

The value of x,y,z & w

$\huge\color{blue}{Solution :-}$

We are given a quadrilateral whose 3 angles are,

60°

80°

120°

We know,

In a quadrilateral angles add up to 360°

In a linear pair the Angles add up to 180°

━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\color{orange}\underline{\bigstar\:\textsf{According to the given Question :}}$

On the way to finding w:-

$\color{green}:\implies \displaystyle\sf 60^{\circ}+80^{\circ}+120^{\circ}+w' = 360^{\circ}:⟹60 </p><p>∘+80∘+120∘+w′=360∘$

$\color{gold}:\implies \displaystyle\sf 260 + w' = 360^{\circ}:⟹260+w ′=360 ∘$

$\color{blue}:\implies \displaystyle\sf w' = 360-260:⟹w ′=360−260$

$\color{olive}:\implies \displaystyle\sf w' = 100^{\circ}:⟹w ′ =100 ∘$

$\huge \color{plum}So \: then \: w \: will \: be,$

$:\implies \displaystyle\sf w+w' = 180^{\circ} \:\:\{ Linear \ Pair\}:⟹w+w ′=180 ∘$

$\color{purple}:\implies \displaystyle\sf w+100 = 180^{\circ}:⟹w+100=180 ∘$

$\color{navy}:\implies \displaystyle\sf w = 180-100:⟹w=180−100$

$\color{cyan}\implies{\underline{\boxed{\frak{\red{w= 80^{\circ}}}}}}\;\bigstar⟹ w=80 ∘$

$\color{teal} ★Similarly \: x \: will \: \: be,$

$\color{teal}:\implies \displaystyle\sf x'+x = 180^{\circ}\:\:\{ Linear \ Pair\}:⟹x ′+x=180 ∘{LinearPair}\color{aqua}:\implies \displaystyle\sf 120^{\circ}+x = 180^{\circ}:⟹120 ∘ +x=180 ∘</p><p></p><p> </p><p></p><p>[tex] \color{orange}:\implies \displaystyle\sf x = 180-120:⟹x=180−120$

$\implies{\underline{\boxed{\frak{\red{x= 60^{\circ}}}}}}\;\bigstar⟹ x=60 ∘$

$\color{yellow}★Value \: of \: y \: will \: be,$

$\color{springgreen}:\implies \displaystyle\sf y'+y = 180^{\circ} \:\:\{ Linear \ Pair\}:⟹y ′+y=180 ∘ {Linear Pair}$

$\color{red}:\implies \displaystyle\sf 80^{\circ}+y = 180^{\circ}:⟹80 ∘+y=180 ∘$

$\color{plum}:\implies \displaystyle\sf y = 180-80:⟹y=180−80$

$\color{lightblue}\implies{\underline{\boxed{\frak{\blue{y= 100^{\circ}}}}}}\;\bigstar⟹ y=100 ∘$

★

★And finally the value of z will be,

$\color{magenta}:\implies \displaystyle\sf z'+z = 180^{\circ}\:\:\{ Linear \ Pair\}:⟹z′+z=180 ∘{Linear Pair}$

$\color{skyblue}:\implies \displaystyle\sf 60^{\circ}+z = 180^{\circ}:⟹60∘+z=180 ∘$

$\color{indigo}:\implies \displaystyle\sf z = 180-60:⟹z=180−60$

$\implies{\underline{\boxed{\frak{\green{z= 120^{\circ}}}}}}\;\bigstar⟹ z=120 ∘$

★

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