Question

help me with the following question​

Answer 1

Angle 1 = Angle 3 ( Vertically opposite Angles)

So , 3x + 15° = x + 5y°

=} 2x -5y +15° = 0 -(1)

Now , AB || CD ( Given )

So , Angle 1 = Angle 5 [ Corresponding Angles ]

=} 3x+ 15° = 7y + 2°

=} 3x -7y + 13° = 0 -(2)

Multiplying eqñ (1) by 3 and eqñ (2) by 2 , we get

6x - 15y +45 = 0 -(3)

6x -14y + 26 = 0 -(4)

Now , Subtracting eqñ (4) from eqñ(3)

6x-15y+45-6x+14y -26 = 0

-y = -19

So , y = 19

Putting this value if y in eqñ(1)

2x - 5 x 19 +15 = 0

2x -95 +15 = 0

2x = 80

x = 40

Now putting these values of x and y in Angle 5's value :

7x19 +2 = 135 °

So , Angle 6 = 180° - Angle 5 [ Linear Pair]

Angle 6 = 180° - 135 ° =} 45°

Solved ...