Measures of three angles of a …

By angle sum property of triangle, the sum of all angles is ${180}^0$, So

(5x + 10) + (3x + 15) + (x + 20) = ${180}^0$

5x + 10 + 3x + 15 + x + 20 = ${180}^0$

9x + 45 = ${180}^0$

9x = ${180}^0$ - ${45}^0$

9x = ${135}^0$

x = $\frac{135}{9} = {15}^0$

Therefore, First angle = 5x + 10 =(5 × 15) + 10 = 75 + 10 = ${85}^0$

                Second angle = 3x + 15 = (3 × 15) + 15 = 45 + 15 = ${60}^0$

                Third angle = x + 20 = 15 + 20 = ${35}^0$

Since, all the angles of the triangle measures less than ${90}^0$, So it is an acute-angled triangle. 

Ans. $We Know, Sum of angles of ∆ is 180°. So$

$5x +10 + 3x+15+x+20 = 180$

=> $9x + 45 = 180$

=> $9x = 135$

=> $x = 15$

$First angle = 5\times 15 + 10 = 85$

$Second Angle = 3 \times 15 + 15 = 60$

$3rd angle = 15 + 20 = 35$

$This is an Acute-angled Triangle$

We will use angle sum property of triangles.  The sum of all the angles of a triangle should be 180 degrees. 

(5x+10)+(3x+15)+(x+20) =180

9x+45=180

9x=135

x=15

So, 5x+10 = 75+10 =85 degrees

3x+15=45+15 = 60 degrees

x+20 = 15+20 =35 degrees

Since all the angles of this triangle measure less than 90 degrees,  therefore it is an Acute-angled triangle.