6. The observation deck of the Skylon Tower in Niagara Falls, Ontario, is A $166 \mathrm{~m}$ above the Niagara River. A tourist in the observation deck notices two boats on the water. From the tourist's position,
- the bearing of boat $A$ is $180^{\circ}$ at an angle of depression of $40^{\circ}$
- the bearing of boat $B$ is $250^{\circ}$ at an angle of depression of $34^{\circ}$
Calculate the distance between the two boats to the nearest metre.

Step 1 ：Let T be the position of the tourist, A be the position of boat A, and B be the position of boat B.

Step 2 ：Let TA = x, TB = y, and AB = z.

Step 3 ：Since the angle of depression of boat A is 40 degrees, we have \(\tan(40^\circ) = \frac{166}{x}\), so \(x = \frac{166}{\tan(40^\circ)}\).

Step 4 ：Similarly, since the angle of depression of boat B is 34 degrees, we have \(\tan(34^\circ) = \frac{166}{y}\), so \(y = \frac{166}{\tan(34^\circ)}\).

Step 5 ：Now, we can use the Law of Cosines on triangle ATB to find the distance between the two boats: \(z^2 = x^2 + y^2 - 2xy\cos(70^\circ)\).

Step 6 ：Substitute the values of x and y: \(z^2 = \left(\frac{166}{\tan(40^\circ)}\right)^2 + \left(\frac{166}{\tan(34^\circ)}\right)^2 - 2\left(\frac{166}{\tan(40^\circ)}\right)\left(\frac{166}{\tan(34^\circ)}\right)\cos(70^\circ)\).

Step 7 ：Calculate z: \(z = \sqrt{\left(\frac{166}{\tan(40^\circ)}\right)^2 + \left(\frac{166}{\tan(34^\circ)}\right)^2 - 2\left(\frac{166}{\tan(40^\circ)}\right)\left(\frac{166}{\tan(34^\circ)}\right)\cos(70^\circ)}\approx 402.6\).

Step 8 ：\(\boxed{403}\) meters is the distance between the two boats to the nearest meter.