Question

❍ The angles of the Quadrilateral are x°, (x – 10)°, (x + 30)° and 2x° respectively.

⠀⠀⠀$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$⠀⠀⠀⠀

The sum of all angles of the Quadrilateral is 360°. Therefore,

$:\implies\sf x + (x - 10)^\circ + (x + 30)^\circ + 2x = 360^\circ \\\\\\:\implies\sf x + x + x + 2x - 10 + 30 = 360^\circ \\\\\\:\implies\sf 3x + 2x - 10 + 30= 360^\circ \\\\\\:\implies\sf 5x + 20 = 360^\circ \\\\\\:\implies\sf 5x = 360^\circ - 20 \\\\\\:\implies\sf 5x = 340^\circ \\\\\\:\implies\sf x = \cancel\dfrac{340^\circ}{5} \\\\\\:\implies{\underline{\boxed{\frak{\purple{x = 68^\circ}}}}}\;\bigstar$

Hence,

First angle, x = 68°
Second angle, (x - 10)° = (68 - 10)° = 58°
Third angle, (x + 30)° = (68 + 30)° = 98°
Fourth angle, 2x = 2(68)° = 136°

$\therefore{\underline{\sf{Hence, the\;greatest\;angle\;is\;\bf{136^\circ }.}}}$

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⠀
V E R I F I C A T I O N :

As we know that sum of the all angles of Quadrilateral is 360°. And, we've measure of each angle. So, Let's verify :

$\dashrightarrow\sf a + b + c + d = 360^\circ \\\\\\\dashrightarrow\sf 68^\circ + 58^\circ + 98^\circ + 136^\circ = 360^\circ \\\\\\\dashrightarrow{\boxed{\underline{\sf{360^\circ = 360^\circ}}}}$

$\therefore{\underline{\textsf{\textbf{Hence Verified!}}}}$
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Answer 1

Explanation:

❍ The angles of the Quadrilateral are x°, (x – 10)°, (x + 30)° and 2x° respectively.

⠀⠀⠀\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}

†Asweknowthat:

⠀⠀⠀⠀

The sum of all angles of the Quadrilateral is 360°. Therefore,

\begin{gathered}:\implies\sf x + (x - 10)^\circ + (x + 30)^\circ + 2x = 360^\circ \\\\\\:\implies\sf x + x + x + 2x - 10 + 30 = 360^\circ \\\\\\:\implies\sf 3x + 2x - 10 + 30= 360^\circ \\\\\\:\implies\sf 5x + 20 = 360^\circ \\\\\\:\implies\sf 5x = 360^\circ - 20 \\\\\\:\implies\sf 5x = 340^\circ \\\\\\:\implies\sf x = \cancel\dfrac{340^\circ}{5} \\\\\\:\implies{\underline{\boxed{\frak{\purple{x = 68^\circ}}}}}\;\bigstar\end{gathered}

:⟹x+(x−10)

∘

+(x+30)

∘

+2x=360

∘

:⟹x+x+x+2x−10+30=360

∘

:⟹3x+2x−10+30=360

∘

:⟹5x+20=360

∘

:⟹5x=360

∘

−20

:⟹5x=340

∘

:⟹x=

5

340

∘

:⟹

x=68

∘

★

Hence,

First angle, x = 68°

Second angle, (x - 10)° = (68 - 10)° = 58°

Third angle, (x + 30)° = (68 + 30)° = 98°

Fourth angle, 2x = 2(68)° = 136°

\therefore{\underline{\sf{Hence, the\;greatest\;angle\;is\;\bf{136^\circ }.}}}∴

Hence,thegreatestangleis136

∘

.

━━━━━━━━━━━━━━━━━━━⠀⠀⠀

⠀

V E R I F I C A T I O N :

As we know that sum of the all angles of Quadrilateral is 360°. And, we've measure of each angle. So, Let's verify :

\begin{gathered}\dashrightarrow\sf a + b + c + d = 360^\circ \\\\\\\dashrightarrow\sf 68^\circ + 58^\circ + 98^\circ + 136^\circ = 360^\circ \\\\\\\dashrightarrow{\boxed{\underline{\sf{360^\circ = 360^\circ}}}}\end{gathered}

⇢a+b+c+d=360

∘

⇢68

∘

+58

∘

+98

∘

+136

∘

=360

∘

⇢

360

∘

=360

∘

${}^{ \please \: mark \: me \: brainlist \: ans.}$

Answer 2

Answer:

❍ The angles of the Quadrilateral are x°, (x – 10)°, (x + 30)° and 2x° respectively.

⠀⠀⠀\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}

†Asweknowthat:

⠀⠀⠀⠀

The sum of all angles of the Quadrilateral is 360°. Therefore,

\begin{gathered}:\implies\sf x + (x - 10)^\circ + (x + 30)^\circ + 2x = 360^\circ \\\\\\:\implies\sf x + x + x + 2x - 10 + 30 = 360^\circ \\\\\\:\implies\sf 3x + 2x - 10 + 30= 360^\circ \\\\\\:\implies\sf 5x + 20 = 360^\circ \\\\\\:\implies\sf 5x = 360^\circ - 20 \\\\\\:\implies\sf 5x = 340^\circ \\\\\\:\implies\sf x = \cancel\dfrac{340^\circ}{5} \\\\\\:\implies{\underline{\boxed{\frak{\purple{x = 68^\circ}}}}}\;\bigstar\end{gathered}

:⟹x+(x−10)

∘

+(x+30)

∘

+2x=360

∘

:⟹x+x+x+2x−10+30=360

∘

:⟹3x+2x−10+30=360

∘

:⟹5x+20=360

∘

:⟹5x=360

∘

−20

:⟹5x=340

∘

:⟹x=

5

340

∘

:⟹

x=68

∘

★