Question

The angles of a triangle are 5(x-1)°,3(2x-5)° and 9x.Find the value of all the angles?​

Answer 1

$\bf{\underline{Question:-}}$

The angles of a triangle are 5(x-1)°,3(2x-5)° and 9x.Find the value of all the angles?

$\bf{\underline{Given:-}}$

• angles of ∆ are
• 5 ( x - 1 )°
• 3 ( 2x - 5 )°
• 9x

$\bf{\underline{To\:Find:-}}$

• Value of the angles

$\bf{\underline{Triangle\: identity:-}}$

• Sum Of Three Angles Of ∆ is = 180°

$\bf{\underline{Solution:-}}$

$\bf → 5(x - 1) + 3(2x - 5 )+9x = 180$

$\bf → 5x - 5 + 6x - 15 + 9x = 180$

$\bf → 5x + 6x + 9x - 5 - 15= 180$

$\bf → 20x -20 = 180$

$\bf → 20x = 180+20$

$\bf → 20x = 200$

$\bf → x = 200/20$

$\bf\large{\underline{\red{x = 10°}}}$

$\bf{\underline{Therefore:-}}$

→ x = 10°

• Now finding the value of given angles

→ First angle 5(x - 1) = 5 ( 10 - 1 ) = 5 × 9 = 45°

→ 2nd angle = 3(2x - 5)

→ 3 ( 2 × 10 - 5)

→ 3( 20 - 5) = 45°

→ third angle = 9x

→ 9 × 10 = 90°

$\bf{\underline{Hence:-}}$

The angles are

• 45°, 45° and 90° respectively