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 = θ radians, a) show that, to 3 decimal places, θ = 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 shown 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 Applications of Differentiation: 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 1 - 2013 - Paper 5

Question 1

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 = θ radians, a) show that, to 3 decimal places,... 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 1 - 2013 - Paper 5

Step 1

show that, to 3 decimal places, θ = 0.865

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Answer

To find the angle θ, we can use the cosine rule, which states that for a triangle with sides a, b, and c, and angle C opposite side c:

$c^2 = a^2 + b^2 - 2ab \cos(C)$

In our case, let:

$a = 7$ m (AB)
$b = 13$ m (AC)
$c = 10$ m (BC)

Substituting the values, we have:

$10^2 = 7^2 + 13^2 - 2 \times 7 \times 13 \cos(θ)$

Calculating $10^2$ , $7^2$ , and $13^2$ gives:

$100 = 49 + 169 - 182 \cos(θ)$

This simplifies to:

$100 = 218 - 182 \cos(θ)$

Rearranging gives:

$182 \cos(θ) = 218 - 100$

$182 \cos(θ) = 118$

Thus:

$\cos(θ) = \frac{118}{182} \approx 0.6484$

Using a calculator to find θ, we get:

$θ \approx \cos^{-1}(0.6484) \approx 0.865 \text{ radians (to 3 decimal places)}$

Step 2

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

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Answer

To find the area of the shaded region S, we first need to calculate the area of triangle ABC using the formula:

$\text{Area} = \frac{1}{2} \times b \times h$

Where:

Base (b) = AC = 13 m
Height (h) = 7 m (as BD is perpendicular to AC)

Thus:

$\text{Area}_{ABC} = \frac{1}{2} \times 13 \times 7 \times \sin(0.865)$

Calculating this gives:

$\text{Area}_{ABC} \approx \frac{1}{2} \times 13 \times 7 \times 0.865 = 38.845$

Next, we find the area of sector ABD:

$\text{Area}_{ABD} = \frac{1}{2} \times r^2 \times θ$

Where:

$r = 7$ m (radius)

Calculating:

$\text{Area}_{ABD} = \frac{1}{2} \times 7^2 \times 0.865 = 21.22$

Now, the area of the shaded region S is:

$\text{Area}_{S} = \text{Area}_{ABC} - \text{Area}_{ABD}$

Calculating gives:

$\text{Area}_{S} = 38.845 - 21.22 = 17.625 ext{ m}^2$

Next, we convert the area into the amount of grass seed required. Since 50 g of grass seed is needed per square metre:

$\text{Amount of seed} = 17.625 \times 50 = 881.25 ext{ g}$

Rounding to the nearest 10 g gives:

$\approx 880 ext{ g}$

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 1 - 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 1 - 2013 - Paper 5

show that, to 3 decimal places, θ = 0.865

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

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