Question

R s agarwal two angles of a triangle are equal and the third angle is greater than each one of them by 18

Answer 1

$\mathbf{Let \: equal \: angles \: are \: \: x° \: \: each, } \\ \mathbf{ So, other \: unequal \: angle = ( x + 18 )°} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (according \: to \: the \: question)$




Sum of all angles of a triangle = 180



So,






=> x + x + x + 18 = 180°

$= > 3x = 180 - 18 \\ \\ = > 3x = 162 \\ \\ = > x = \frac{162}{3} \\ \\ = > x = 54$




Hence, Angles are




$\mathbf{ Equal \: angles = 54° each} \: \\ \mathbf{ Unequal = ( 54 + 18 ) = 72° }$

Answer 2

X+X+X+18=180°
=>3X=180°-18=162
=>X=54°

THE ANGLE IS (54°+18)=72°