A vertical mobile phone mast, [DC], of height h m, is secured with two cables: [AC] of length x m, and [BC] of length y m, as shown in the diagram - Leaving Cert Mathematics - Question 7 - 2020

Using the sine rule in triangle ABC:

$$\frac{y}{\sin 30°} = \frac{100}{\sin 105°}.$$

Solving for y:

$$y = 100 \times \frac{\sin 30°}{\sin 105°} = 100 \times \frac{0.5}{0.9659} ≈ 51.8 m.$$

Thus, the length of y is approximately 51.8 m.

Using triangle ABC again:

For h:

Using the sine rule:

$$\frac{h}{\sin 45°} = \frac{y}{\sin 30°} = \frac{51.8}{0.5}.$$

This gives:

$$h = \frac{51.8 \times \sin 45°}{\sin 30°} = \frac{51.8 \times 0.7071}{0.5} ≈ 73.6 m.$$

For x:

Using the sine rule:

$$\frac{x}{\sin 30°} = \frac{h}{\sin 45°}.$$

Substituting h:

$$x = \frac{h \times \sin 30°}{\sin 45°} = \frac{73.6 \times 0.5}{0.7071} ≈ 52.2 m.$$

Thus, the height h is approximately 73.6 m and x is approximately 52.2 m.

Cost of cables = 2 \times 25 \times x m\n Cost of mast = 580 Euros. Total cost before VAT:

$$Total = (2 \times 25and \times 73.2) + 580 = 29,584.19 Euros.$$

Including VAT of 23%:

$$TotalCost = TotalCost \times 1.23 = 29,584.19 \times 1.23 = 35,172.10 Euros.$$

Area of a regular hexagon with side length s is given by:

$$Area = \frac{3\sqrt{3}}{2} s^2.$$

For s = 8 km:

$$Area = \frac{3\sqrt{3}}{2} (8)^2 = 166.28 km².$$

Thus, the area is approximately 166.28 km².

The area of the circle inscribed in the hexagon is given by:

Area = \frac{3\sqrt{3}}{2}}s^2 - \text{Area of hexagon}.

This calculation results in an area of approximately 5.8 km².