Question

11. (3) since, ( A B C D ) is a cyclic quadrilateral.
( therefore )
( angle A+angle C=180^{circ} )
( Rightarrow 2 x+4+4 y-4=180^{circ} )
( Rightarrow quad x+2 y=90^{circ} )
Also, ( angle B+angle D=180^{circ} )
( Rightarrow 5 y+5+x+10=180^{circ} )
( Rightarrow quad x+5 y=165^{circ} )
On subtracting Eq.
(i) from Eq.
(ii), we get
( 3 y=75^{circ} )
( Rightarrow )
( y=25^{circ} )

# In the following figure, find the 11. In the for value of y. А (2x+4) DK«x+10) (5y+5) B (4y-4) (1) 20" (2) 10° (3) 25° (4) 30°

Solution