Given that sin θ = 4/5 and tan θ < 0. Determine with the aid of a diagram and WITHOUT the use of a calculator: 5.1.1 tan θ 5.1.2 2cos²θ − 1

In right-angled triangle ABC:

$$\tan(15^\circ) = \frac{BC}{AB}$$ $$BC = AB \times \tan(15^\circ)$$

Substitute the given value for AB:

$$BC = 30 \times \tan(15^\circ)$$

In right-angled triangle ABD:

$$\tan(21^\circ) = \frac{BD}{AB}$$ $$BD = AB \times \tan(21^\circ)$$

Substitute the given value for AB:

$$BD = 30 \times \tan(21^\circ)$$

The distance between the two windows is the difference between BD and BC.

$$CD = BD - BC$$ $$CD = (30 \times \tan(21^\circ)) - (30 \times \tan(15^\circ))$$

Factor out 30:

$$CD = 30 (\tan(21^\circ) - \tan(15^\circ))$$

Using a calculator:

$$\tan(21^\circ) \approx 0.38386$$ $$\tan(15^\circ) \approx 0.26795$$ $$CD = 30 (0.38386 - 0.26795)$$ $$CD = 30 (0.11591)$$ $$CD \approx 3.4773$$

Rounding to two decimal places, as is common for lengths in such problems:

$$CD \approx 3.48 \text{ m}$$

The distance between the two windows is 3.48 m.

$$3.48 \text{ m}$$