From a point A, the angle of elevation of the top of a watch tower is 20°. From another point B which is 80m from the base of the tower, the angle of elevation of the top of the tower is 26°. Determin

Let $h$ be the height of the tower.
Let $F$ be the base of the tower and $T$ be the top of the tower.
Point $B$ is $80 \, \text{m}$ from the base of the tower, so $FB = 80 \, \text{m}$.
The angle of elevation from $B$ to the top of the tower is $26^\circ$.
In the right-angled triangle $\triangle TFB$:

$$\tan(\angle TBF) = \frac{TF}{FB}$$ $$\tan(26^\circ) = \frac{h}{80}$$

$$h = 80 \times \tan(26^\circ)$$ $$h = 80 \times 0.487725889$$ $$h \approx 39.01807 \, \text{m}$$

Rounding to three decimal places, the height of the tower is:

$$\boxed{39.018 \, \text{m}}$$

The angle of elevation from $A$ to the top of the tower is $20^\circ$.
Let $FA$ be the distance from point $A$ to the base of the tower.
In the right-angled triangle $\triangle TFA$:

$$\tan(\angle TAF) = \frac{TF}{FA}$$ $$\tan(20^\circ) = \frac{h}{FA}$$

$$FA = \frac{h}{\tan(20^\circ)}$$

Using the unrounded value of $h \approx 39.01807112 \, \text{m}$:

$$FA = \frac{39.01807112}{0.363970234}$$ $$FA \approx 107.20690 \, \text{m}$$

Since both points A and B are on the same side of the tower, and B is closer to the tower (as its angle of elevation is larger), the distance between A and B is $FA - FB$.

$$d_{AB} = FA - FB$$ $$d_{AB} = 107.20690 - 80$$ $$d_{AB} = 27.20690 \, \text{m}$$

Rounding to three decimal places, the distance between A and B is:

$$\boxed{27.207 \, \text{m}}$$