Question

Can someone help me with this please?

Answer 1

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