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.

A-Level Edexcel Maths Pure Basic Trigonometry: 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 9 - 2013 - Paper 5

Question 9

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 9 - 2013 - Paper 5

Step 1

show that, to 3 decimal places, θ = 0.865

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Answer

To determine the angle θ, we can utilize the cosine rule:

$ext{cos } θ = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC}$

Substituting the given lengths:

AB = 7 m
AC = 13 m
BC = 10 m

We perform the calculation:

$\text{cos } θ = \frac{7^2 + 13^2 - 10^2}{2 \times 7 \times 13} = \frac{49 + 169 - 100}{182} = \frac{118}{182} = \frac{59}{91} \approx 0.6494$

Using the calculator, we find:

$θ = \text{cos}^{-1}(0.6494) \approx 0.865 \text{ radians}$

Thus, θ to three decimal places is 0.865.

Step 2

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

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Answer

First, we must calculate the area of the shaded region S. We need to find the areas of triangle ABC and sector ABD.

Area of Triangle ABC:

We use the formula for the area of a triangle:

$Area_{ABC} = \frac{1}{2} \times AB \times AC \times \sin(θ)$

Substituting the values, we have:

$Area_{ABC} = \frac{1}{2} \times 7 \times 13 \times \sin(0.865) = \frac{1}{2} \times 7 \times 13 \times 0.7651 = 21.7 m^2$
Area of Sector ABD:

The formula for the area of a sector is:

$Area_{sector} = \frac{1}{2} r^2 θ$

Here, r = 7 m and θ = 0.865 radians:

$Area_{sector - ABD} = \frac{1}{2} \times 7^2 \times 0.865 = 21.2 m^2$
Area of Shaded Region S:

Therefore, the area of the shaded region S is calculated as:

$Area_{S} = Area_{ABC} - Area_{sector - ABD} = 21.7 - 21.2 = 0.5 m^2$

Since 50 g of grass seed are needed for each square metre of lawn, the total amount of seed required is:

$Amount_{seed} = 0.5 \times 50 g = 25 g$

Rounding to the nearest 10 g, we find the answer is 30 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 9 - 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 9 - 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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