Question

please answer question no 10

Answer 1

in abcd,

3x+4+50+3x+10+y+5x+8+x=360 (bdc=y)

then 12x+y=288 ....1

now in∆bcd,

5x+8+x+y=180

6x+y=172 .....2

eq 2-1,

12x+y-(6x+y)=288-172

6x=116

x=116÷6

(b)adb= 3x+10

=3×116÷6+10

=68