Question

If three angles of a triangle are 2x+10 ,3x-10 and 4x-18.the smallest angle will be?

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Answer 1

$\large{\red{\underline{\underline{\bold{Given:-}}}}}$

• If three angles of triangle are 2x+10, 3x-10, 4x-18 are given

$\large{\blue{\underline{\underline{\bold{To find}}}}}$

• The smallest angle of a triangle if three angles of triangle are given as 2x+10, 3x-10, 4x-18

$\large{\green{\underline{\underline{\bold{Solution}}}}}$

$\rightarrow$ Three angles of triangle are 2x+10, 3x-10, 4x-18

$\rightarrow$ Sum of three angles of triangle= 180°

$\rightarrow$ So 2x+10 + 3x-10 + 4x-18= 180°

$\rightarrow$ 9x - 18= 180°

$\rightarrow$ 9x= 180° + 18

$\rightarrow$ 9x= 198°

$\rightarrow$ x=22°

• Now substitute value of x in each angle to get smallest angle

$\longrightarrow$ 1st angle =2x+10 = 2×22+10=54

$\longrightarrow$ 2nd angle =3x-10= 3×22-10=56

$\longrightarrow$ 3rd angle = 4x-18= 4×22-18=70

• So angle 1st is smallest angle

$\longrightarrow$ angle 1st=54° is smallest one

Answer 2

Answer:

Given - angles of a triangle

1st angle =2x+10 ,

2nd angle = 3x-10

3rd angle =4x-18

To find - smallest angle of a triangle

Solution -

Sum of all angle of triangle = 180°

2x+ 10 + 3x-10 + 4x-18 = 180

2x+ 3x+ 4x+ 10-10-18= 180

9x- 18= 180

9x= 180+18

9x= 198

x= 22

2x+10= 2(22)+10= 44+10= 54°

3x-10= 3(22)+10= 66+10= 76°

4x- 18= 4(22)-18= 88-18= 90 °

Therefore, smallest angle = 54°