Math, asked by prabhatmaudha, 4 months ago

find the value of x in this question
please give the correct answer​

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Answers

Answered by belwaldeepak123
2

x+3x+5x+4x+5x=360°

[as they will add to form a complete angle]

18x=360°

x=360°÷18

x= 20°

Answered by snehitha2
4

Answer :

The required value of x is 20°

Step-by-step explanation :

Concept : The sum of the angles around a point is equal to 360°

    \setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\put(0,0){\vector(1,0){2}}\put(0,0){\vector(2,1){2}}\put(0,0){\vector(-2,1){2}}\put(0,0){\vector(2,-1){2}}\put(0,0){\vector(-2,-1){2}}\qbezier(-0.25,0.2)(0,0.5)(0.2,0.1)\qbezier(0.4,0.2)(0.6,0.2)(0.4,0)\qbezier(0.5,0)(0.6,-0.2)(0.5,-0.2)\qbezier(0.3,-0.1)(0,-0.7)(-0.3,-0.2)\qbezier(-0.4,-0.3)(-0.7,0.2)(-0.3,0.2)\put(0,0.6){5x}\put(0.7,0.1){x}\put(0.6,-0.3){3x}\put(0,-0.8){5x}\put(-0.9,0){4x}\end{picture}

Given angles around a point : 5x , x , 3x , 4x , 5x

   Their sum = 360°

5x + x + 3x + 4x + 5x = 360°

  18x = 360°

    x = 360°/18

    x = 20°

The value of x is 20°

_______________________

  • Angles around a point add up to 360°
  • Angles on a straight line add up to 180°

       \setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\put(5,1){\vector(1,0){4}}\put(5,1){\vector(-1,0){4}}\put(5,1){\vector(1,1){3}}\put(2,2){$\underline{\boxed{\large\sf a + b = 180^{\circ}}$}}\put(4.5,1.3){$\sf a^{\circ}$}\put(5.7,1.3){$\sf b^{\circ}$}\end{picture}

  • Alternate angles are equal.

       \setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(0,3){2}{\line(1,0){4}}\qbezier(0,0)(0,0)(4,3)\qbezier(1,0)(1.2,0.35)(0.8,0.6)\qbezier(3,3)(2.8,2.65)(3.2,2.4)\put(2.5,0.02){\vector(1,0){0}}\put(1.5,3.02){\vector(-1,0){0}}\end{picture}

  • Corresponding angles are equal.

        \setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(2,5)\put(2,5){\line(1,0){4}}\put(1,2.5){\line(1,0){4}}\qbezier(1.8,2.5)(1.8,1.85)(0.75,1.88)\qbezier(2.8,5)(2.8,4.3)(1.75,4.3)\put(4,5){\vector(1,0){0}}\put(3.5,2.52){\vector(1,0){0}}\end{picture}

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