Question

In a traingle the angles are 2×,(×+30),(×-10) find the angles ​

Answer 1

Appropriate Question :

In a traingle the angles are 2x , (x+30), (x-10). Find the angles.

$\Large {\underline { \sf {Clarification :}}}$

Here, we are given that the angles of a triangle are 2x , (x+30) , (x-10). We need to find the value of the angles.

In order to find the measure of the angles, we need to calculate the value of x. For that, we'll form a linear equation. As we know that, the sum of all the interior angles of a triangle is equivalent to 180°. So, in L.H.S of the equation we'll perform addition of the angles and in R.H.S we'll write 180°. By using Transposition method, we'll find the value of x . After that, we'll substitute those values in the expressions of the angles we are given here to find the measure of the angles.

$\bf \red { \dag }$ Transposition method :

• This is the method used to solve a linear equation having variables and constants.
• In this method, we transpose the values from R.H.S to L.H.S and vice-versa and changes its sign while transposing to find the value of the unknown value.

$\Large {\underline { \sf {Explication \: of \: steps :}}}$

$\underline{ \pmb { \mathit { Given \; Information :}}}$

• First angle = 2x°

• Second angle = (x + 30)°

• Third angle = (x - 10)°

$\underline{ \pmb { \mathit {To \; calculate :}}}$

• Measure of the angles.

$\underline{ \pmb { \mathit { Calculation :}}}$

As we know that,

❝ Angle sum property of triangle states that the sum of all interior angles of a triangle is 180°. ❞

$\longmapsto\underline{ \boxed {\bf {Sum \: of \: angles_{(\Delta)} = 180^\circ }}}$

Substituting the values,

$\longrightarrow \sf { 2x^\circ + (x + 30)^\circ + (x -10)^\circ = 180^\circ }$

Remove the brackets.

$\longrightarrow \sf { 2x^\circ + x^\circ + 30^\circ + x^\circ -10^\circ = 180^\circ }$

Grouping all like terms.

$\longrightarrow \sf { 2x^\circ + x^\circ + x^\circ + 30^\circ -10^\circ = 180^\circ }$

Performing addition and subtraction in L.H.S.

$\longrightarrow \sf { 4x^\circ + 20^\circ = 180^\circ }$

Transposing 20 from L.H.S to R.H.S and changing its sign.

$\longrightarrow \sf { 4x^\circ = 180^\circ - 20^\circ }$

Performing subtraction in R.H.S.

$\longrightarrow \sf { 4x^\circ = 160^\circ }$

Now, transposing 4 from L.H.S to R.H.S, changing its sign.

$\longrightarrow \sf { x^\circ =\cancel{ \dfrac{160^\circ}{4} }}$

$\longrightarrow \underline {\boxed{\sf { x^\circ = 40^\circ }}} \: \: \red { \bigstar }$

Therefore, value of x is 40°.

$\underline{\bf {According\; to \;the\; question,}}$

$\dashrightarrow \underline {\sf { \dag \; First \; Angle \; of \; \Delta }}$

$\longrightarrow \sf {First \: angle = 2x^\circ }$

$\longrightarrow \sf {First \: angle = 2(40)^\circ }$

$\longrightarrow \sf \red {First \: angle = 80^\circ }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow \underline {\sf { \dag \; Second \; Angle \; of \; \Delta }}$

$\longrightarrow \sf {Second \: angle = (x+30)^\circ }$

$\longrightarrow \sf {Second \: angle = (40+30)^\circ }$

$\longrightarrow \sf \red {Second \: angle = 70^\circ }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow \underline {\sf { \dag \; Third \; Angle \; of \; \Delta }}$

$\longrightarrow \sf {Third \: angle = (x-10)^\circ }$

$\longrightarrow \sf {Third \: angle = (40-10)^\circ }$

$\longrightarrow \sf \red {Third \: angle = 30^\circ }$

Therefore, the angles of the triangle are 80° , 70° and 30°.