Figure 2 shows the design for a triangular garden ABC where AB = 7 m, AC = 13 m and BC = 10 m. Given that angle BAC = $ \theta $ radians, (a) show that, to 3 decimal places, $ \theta = 0.865 $ (3) The point D lies on AC such that BD is an arc of the circle centre A, radius 7 m. The shaded region S is bounded by the arc BD and the lines BC and DC. The shaded region S will be seen with grass seed, to make a lawned area. Given that 50 g of grass seed are needed for each square metre of lawn, (b) find the amount of grass seed needed, giving your answer to the nearest 10 g. (7)

A-Level Edexcel Maths Pure Modelling with Functions: Figure 2 shows the design for a

Figure 2 shows the design for a triangular garden ABC where AB = 7 m, AC = 13 m and BC = 10 m - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 5

Question 8

Figure 2 shows the design for a triangular garden ABC where AB = 7 m, AC = 13 m and BC = 10 m. Given that angle BAC = $ \theta $ radians, (a) show that, to 3 dec... show full transcript

Worked Solution & Example Answer:Figure 2 shows the design for a triangular garden ABC where AB = 7 m, AC = 13 m and BC = 10 m - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 5

Step 1

Given that angle BAC = $ \theta $ radians, show that, to 3 decimal places, $ \theta = 0.865 $

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Answer

To find ( \theta ), we can use the cosine rule in triangle ABC:

\cos \theta = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC}

Substituting the values,

\cos \theta = \frac{7^2 + 13^2 - 10^2}{2 \times 7 \times 13} = \frac{49 + 169 - 100}{182} = \frac{118}{182} = \frac{59}{91}

Now, to find ( \theta ), we take the inverse cosine:

\theta = \cos^{-1}\left(\frac{59}{91}\right) \approx 0.865 \text{ (to 3 decimal places)}

Step 2

find the amount of grass seed needed, giving your answer to the nearest 10 g.

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Answer

First, we find the area of the shaded region S.

Calculate the area of triangle ABC:
$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\theta)$
Substituting the values:
$\text{Area} = \frac{1}{2} \times 7 \times 13 \times \sin(0.865) \approx \frac{1}{2} \times 7 \times 13 \times 0.756 \approx 34.6 \, m^2$
Determine the area of sector ADB:
$\text{Area of Sector} = \frac{1}{2} \times r^2 \times \theta = \frac{1}{2} \times 7^2 \times 0.865 \approx 21.2 \, m^2$
Calculate the area of region S:
$\text{Area of shaded region S} = \text{Area of triangle ABC} - \text{Area of sector ADB}$ $\text{Area of shaded region S} \approx 34.6 - 21.2 = 13.4 \, m^2$
Finally, calculate the amount of grass seed needed:

Since 50 g of seed is needed for each square metre:
$\text{Total Seed} = 13.4 \times 50 = 670 g$
Rounding to the nearest 10 g, the answer is 670 g.

Figure 2 shows the design for a triangular garden ABC where AB = 7 m, AC = 13 m and BC = 10 m - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 5

Worked Solution & Example Answer:Figure 2 shows the design for a triangular garden ABC where AB = 7 m, AC = 13 m and BC = 10 m - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 5

Given that angle BAC = \( \theta \) radians, show that, to 3 decimal places, \( \theta = 0.865 \)

find the amount of grass seed needed, giving your answer to the nearest 10 g.

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