Point A lies at coordinate (4,3) and point B at (6,8). Determine: i) The distance AB ii) The gradient of the straight line AB. iii) The angle AB makes with the horizontal

From part (ii), the gradient $m$ of the straight line AB is:

$$m = \frac{\Delta y}{\Delta x} = \frac{8-3}{6-4} = \frac{5}{2}$$

The gradient $m$ of a line is equal to the tangent of the angle $\theta$ it makes with the positive horizontal axis:

$$m = \tan \theta$$

Substitute the value of $m$:

$$\tan \theta = \frac{5}{2}$$ $$\tan \theta = 2.5$$

To find $\theta$, take the inverse tangent (arctan) of 2.5:

$$\theta = \arctan(2.5)$$

Using a calculator:

$$\theta \approx 68.19859...^\circ$$

Rounding to one decimal place:

$$\theta \approx \boxed{68.2^\circ}$$

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