Answer:
Solution given:
<J=90°[tangent to the circle is perpendicular to radius]
In right angled triangle ∆KJL
base[b]=10
hypotenuse [h]=x+10
perpendicular [p]=24
by using Pythagoras law
p²+b²=h²
24²+10²=(x+10)²
(x+10)²=676
x+10=√676
x=26-10
x=16
