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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

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Question 7

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.png

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. The angle of elev... show full transcript

Worked Solution & Example Answer: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

Step 1

Explain why ∠LBCA is 105°.

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Answer

In triangle ABC, the sum of angles in a triangle equals 180°. Given that ∠A is 30° and ∠B is 45°, we can calculate ∠C as follows:

C=180°(A+B)=180°(30°+45°)=180°75°=105°.∠C = 180° - (∠A + ∠B) = 180° - (30° + 45°) = 180° - 75° = 105°.

Thus, the angle ∠LBCA is 105°.

Step 2

The length of y.

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Answer

Using the sine rule in triangle ABC:

ysin30°=100sin105°.\frac{y}{\sin 30°} = \frac{100}{\sin 105°}.

Solving for y:

y=100×sin30°sin105°=100×0.50.965951.8m.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.

Step 3

Find the value of h and the value of x.

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Answer

Using triangle ABC again:

For h:

Using the sine rule:

hsin45°=ysin30°=51.80.5.\frac{h}{\sin 45°} = \frac{y}{\sin 30°} = \frac{51.8}{0.5}.

This gives:

h=51.8×sin45°sin30°=51.8×0.70710.573.6m.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:

xsin30°=hsin45°.\frac{x}{\sin 30°} = \frac{h}{\sin 45°}.

Substituting h:

x=h×sin30°sin45°=73.6×0.50.707152.2m.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.

Step 4

Calculate the total cost of the cables and mast after VAT is included.

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Answer

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

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

Including VAT of 23%:

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

Step 5

Find the area of the hexagon.

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Answer

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

Area=332s2.Area = \frac{3\sqrt{3}}{2} s^2.

For s = 8 km:

Area=332(8)2=166.28km2.Area = \frac{3\sqrt{3}}{2} (8)^2 = 166.28 km².

Thus, the area is approximately 166.28 km².

Step 6

Find this shaded area.

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Answer

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.

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