An emblem, as shown in Figure 1, consists of a triangle ABC joined to a sector CBD of a circle with radius 4 cm and centre B. The points A, B and D lie on a straight line with AB = 5 cm and BD = 4 cm. Angle BAC = 0.6 radians and AC is the longest side of the triangle ABC. (a) Show that angle ABC = 1.76 radians, correct to 3 significant figures. (b) Find the area of the emblem.

A-Level Edexcel Maths Pure Exponential & Logarithms: An emblem, as shown in Figure 1,

An emblem, as shown in Figure 1, consists of a triangle ABC joined to a sector CBD of a circle with radius 4 cm and centre B - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 4

Question 5

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An emblem, as shown in Figure 1, consists of a triangle ABC joined to a sector CBD of a circle with radius 4 cm and centre B. The points A, B and D lie on a straight... show full transcript

Worked Solution & Example Answer:An emblem, as shown in Figure 1, consists of a triangle ABC joined to a sector CBD of a circle with radius 4 cm and centre B - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 4

Step 1

Show that angle ABC = 1.76 radians, correct to 3 significant figures.

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Answer

To show that angle ABC is 1.76 radians, we can use the sine rule. First, we need to express ABC using the known angles in triangle ABC:

Apply the sine rule: $\frac{\sin(ABC)}{AC} = \frac{\sin(0.6)}{5}$
We know that AC is the longest side, so we can express AC in terms of angle ABC: $AC = b = \sqrt{5^2 + 4^2 - 2 \cdot 5 \cdot 4 \cdot \cos(0.6)}$
Calculate AC: $AC = \sqrt{25 + 16 - 40 \cdot \cos(0.6)} \approx \sqrt{34.8} \approx 5.9$ (using $\, \cos(0.6) \, \approx 0.84$ )
Now applying the sine rule: $\sin(ABC) = \frac{AC \cdot \sin(0.6)}{5}$
Substitute AC into the equation: $\sin(ABC) = \frac{5.9 \cdot \sin(0.6)}{5}$
Solve for angle ABC using the arcsin function: $ABC = \arcsin(\frac{5.9 \cdot 0.564}{5}) \approx \arcsin(0.667) \approx 1.76 \, radians$ (correct to 3 significant figures).

Thus, we have shown that angle ABC = 1.76 radians.

Step 2

Find the area of the emblem.

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Answer

The area of the emblem consists of the area of triangle ABC and the area of sector CBD.

Area of Sector CBD:
The sector area can be calculated using the formula: $\text{Sector area} = \frac{1}{2} r^2 \theta$ where $r = 4$ cm and $\theta = 1.76$ radians.

$\text{Sector area} = \frac{1}{2} \cdot 4^2 \cdot 1.76 = \frac{1}{2} \cdot 16 \cdot 1.76 = 14.08 \, cm^2$
Area of Triangle ABC:
The area can be calculated using the formula: $\text{Area} = \frac{1}{2} \times base \times height$ Using the lengths found previously, we also know one side and height from angle BAC. Therefore we apply: $\text{Area} = \frac{1}{2} \times 5 \times 4 \times \sin(0.6) \approx 5 \times 4 \times 0.564 = 11.28 \, cm^2$
Total Area of the Emblem:
We sum the areas of the sector and triangle: $\text{Total Area} = 14.08 + 11.28 \approx 25.36 \, cm^2$

Thus, the area of the emblem is 25.36 cm².

An emblem, as shown in Figure 1, consists of a triangle ABC joined to a sector CBD of a circle with radius 4 cm and centre B - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 4

Worked Solution & Example Answer:An emblem, as shown in Figure 1, consists of a triangle ABC joined to a sector CBD of a circle with radius 4 cm and centre B - Edexcel - A-Level Maths Pure - Question 5 - 2010 - Paper 4

Show that angle ABC = 1.76 radians, correct to 3 significant figures.

Find the area of the emblem.

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