For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then, the angle of inclination of the inclined plane is

The correct option is 30°.

To find the angle of inclination of the inclined plane, let us analyze the formulas for the maximum range of a projectile projected up and down an inclined plane.

Let:
- $u$ be the velocity of projection,
- $g$ be the acceleration due to gravity, and
- $\beta$ be the angle of inclination of the plane.

The maximum range up the inclined plane for a given velocity $u$ is given by the formula:

$R_{\text{up}}=\frac{u^2}{g(1+\sin \beta )}$

Similarly, the maximum range down the inclined plane for the same velocity $u$ is given by the formula:

$R_{\text{down}}=\frac{u^2}{g(1-\sin \beta )}$

According to the problem statement, the maximum range down the plane is three times the maximum range up the plane:

$R_{\text{down}}=3R_{\text{up}}$

Substituting the expressions for both ranges into the relation, we get:

$\frac{u^2}{g(1-\sin \beta )}=3\cdot \frac{u^2}{g(1+\sin \beta )}$

We can cancel out the common terms $u^2$ and $g$ from both sides of the equation:

$\frac{1}{1-\sin \beta }=\frac{3}{1+\sin \beta }$

Cross-multiplying to solve for $\sin \beta$ gives:

$1+\sin \beta =3(1-\sin \beta )$

Expanding the right side:

$1+\sin \beta =3-3\sin \beta$

Rearranging the terms to group the sine terms together:

$\sin \beta +3\sin \beta =3-1$

$4\sin \beta =2$

$\sin \beta =\frac{2}{4}=\frac{1}{2}$

Since $\sin (30°)=\frac{1}{2}$, we have:

$\beta =30°$

Thus, the angle of inclination of the inclined plane is 30°.