FIGURE I shows a ramp leading to the entrance of a building. B, C and D lie on the same horizontal plane. The perpendicular height (AC) of the ramp is 0.5 m and the angle of elevation from B to A is 15°. The entrance of the building (AE) is 0.915 m wide. FIGURE II shows the top view of the ramp. The area of the top of the ramp is divided into three triangles, as shown in the diagram. If ∠BAE = 120°, calculate the length of BE. 8.3 Calculate the area of ABFD if ∠BFD = 75°; BF = FD and BF = \frac{5}{7} BE.

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FIGURE I shows a ramp leading to the entrance of a building - NSC Mathematics - Question 8 - 2022 - Paper 2

Question 8

FIGURE I shows a ramp leading to the entrance of a building. B, C and D lie on the same horizontal plane. The perpendicular height (AC) of the ramp is 0.5 m and the ... show full transcript

Worked Solution & Example Answer:FIGURE I shows a ramp leading to the entrance of a building - NSC Mathematics - Question 8 - 2022 - Paper 2

Step 1

Calculate the length of AB.

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Answer

Using the sine ratio, we have: $AB = \frac{0.5}{\sin(15^\circ)}$ Calculating this gives: $AB = 1.93 \text{ m}.$

Step 2

If ∠BAE = 120°, calculate the length of BE.

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Answer

To find BE, we can use the cosine rule: $BE^2 = AB^2 + AE^2 - 2(AB)(AE)\cos(BAE).$ Substituting the values: $BE^2 = (1.93)^2 + (0.915)^2 - 2(1.93)(0.915)\cos(120^\circ)$ Calculating this gives: $BE = 2.52 \text{ m}.$

Step 3

Calculate the area of ABFD if ∠BFD = 75°; BF = FD and BF = \frac{5}{7} BE.

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Answer

First, we find BF: $BF = \frac{5}{7} BE = \frac{5}{7}(2.52) = 1.80 \text{ m}.$ Now, we can calculate the area of triangle ABFD: $\text{Area}_{ABFD} = \frac{1}{2}(BF)(FD)\sin(BFD) = \frac{1}{2}(1.80)(1.80)\sin(75^\circ)$ This results in: $\text{Area}_{ABFD} = 1.56 \text{ m}^2.$

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