Question

If 2a°, 3a°,4a° and 6a° are the the angles of a quadrilateral, find them
Please show in full process

Answer 1

Answer:

Sum of angles :- 2a° + 3a° + 4a° +6a° = 360°

15 a° = 360°

a° = 360°/15

a°= 24°

put a°=24° in

2a°=2(24°) =48°

3a°= 3(24°) =72°

4a° = 4(24°) =96°

6a° = 6(24°) =144°

Verify :- 48° + 72° +96° +144° = 360 °

360° = 360°

Hence , this is correct .

Answer 2

$\begin{gathered}\sf \large \red{\underline{ Question:-}}\\\\\end{gathered}$

If 2a°, 3a°,4a° and 6a° are the the angles of a quadrilateral, find them

$\begin{gathered}\\\\\sf \large \red{\underline{Given:-}}\\\\\end{gathered}$

Angles of the quadrilateral are given = 2a°, 3a°,4a° and 6a°

$\begin{gathered}\\\\\sf \large \red{\underline{To \: Find:-}}\\\\\end{gathered}$

The angles of the triangle.

$\begin{gathered}\\\\\sf \large \red{\underline{Solution :- }}\\\\\end{gathered}$

First we need to find the value of a to find the value of the angles of the quadrilateral:–

$\large\underbrace{ \sf { We \: know \: angles \: of \: a \: quadrilateral\: = 360 ^ \circ}}$

$\qquad \quad : \longrightarrow \sf{2a + 3a + 4a + 6a = 360}$

$\qquad \quad : \longrightarrow \sf{15a = 360}$

$\qquad \quad : \longrightarrow \sf{a = \dfrac{ \cancel{360}^{ \large24} }{ \cancel{15}} _ {\large{1}} }$

$\qquad \quad : \leadsto \underline{\boxed{\sf \pink{a = 24}}}$

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Now when we know the value of a let find the angles of the quadrilateral:–

$\qquad \quad \bull \sf \quad{2a = 2 \times 24 = \pink{48 ^ \circ}}$

$\qquad \quad \bull \sf \quad{3a = 3 \times 24 = \pink{72 ^ \circ}}$

$\qquad \quad \bull \sf \quad{4a = 4 \times 24 = \pink{96 ^{ \circ}} }$

$\qquad \quad \bull \sf \quad{6a = 6 \times 24 = \pink{144 ^ \circ}}$

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