To solve for the length of the common external tangent $KL$ between two circles with radii $r_1 = 4$ cm and $r_2 = 2$ cm and given the angle between the radii and the tangent line at the points of tangency is $60^\circ$, we can use the formula for the length of a common external tangent between two circles:

$L = \sqrt{d^2 - (r_1 + r_2)^2}$

where:

• $d$ is the distance between the centers of the circles,
• $r_1$ and $r_2$ are the radii of the circles.

1. Calculate the Distance $d$: Since we have a $60^\circ$ angle between the radii, we can apply the law of cosines in the triangle formed by the centers of the circles and one of the tangent points.

   $d = \sqrt{r_1^2 + r_2^2 - 2 \cdot r_1 \cdot r_2 \cdot \cos(60^\circ)}$

   Substituting the values: $d = \sqrt{4^2 + 2^2 - 2 \cdot 4 \cdot 2 \cdot \cos(60^\circ)}$ $d = \sqrt{16 + 4 - 2 \cdot 4 \cdot 2 \cdot 0.5}$ $d = \sqrt{16 + 4 - 8} = \sqrt{12} = 2\sqrt{3} \, \text{cm}$

2. Find the Length of the Tangent $KL$: Now we apply the external tangent formula:

   $KL = \sqrt{d^2 - (r_1 - r_2)^2}$

   Substitute $d = 2\sqrt{3}$, $r_1 = 4$, and $r_2 = 2$:

   $KL = \sqrt{(2\sqrt{3})^2 - (4 - 2)^2}$ $KL = \sqrt{4 \cdot 3 - 2^2}$ $KL = \sqrt{12 - 4} = \sqrt{8} = 2\sqrt{2} \, \text{cm}$