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### Problem Statement Recap:
You are given a shape \( BCD \), designed for a logo. The curve \( BC \) is an arc of a circle with center \( A \), radius 6 cm, and the angle \( \angle BAC = 2.2 \) radians. Additionally, \( AD = 4 \) cm, and the straight lines \( DB \) and \( DC \) are of equal length.

#### Required Calculations:
(a) Find the area of the sector \( BAC \) in cm².  
(b) Calculate the size of \( \angle DAC \), in radians, to 3 significant figures.  
(c) Determine the complete area of the logo design, to the nearest cm².

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### Let's tackle each part:

#### (a) Area of Sector \( BAC \):
The formula for the area of a sector is:
\[
\text{Area of Sector} = \frac{1}{2} r^2 \theta
\]
Where:
- \( r = 6 \) cm (radius)
- \( \theta = 2.2 \) radians (angle at \( A \))

#### (b) Size of \( \angle DAC \):
This will involve using some trigonometry, as the lines \( DB \) and \( DC \) are equal, and we can use the known values to calculate this.

#### (c) Complete Area of the Logo:
The logo consists of:
- The area of the sector \( BAC \),
- Two triangles \( ABD \) and \( ACD \), which can be calculated using the area formula for triangles with the given dimensions and angles.

I'll start by calculating these values.### (a) The area of the sector \( BAC \) is \( 39.6 \, \text{cm}^2 \).

Now, let's proceed with calculating the size of \( \angle DAC \) for part (b). Since we know that \( DB = DC \), we can use the cosine rule or other trigonometric relations to compute this angle. I'll calculate that next.For part (b), to calculate the size of \( \angle DAC \), we can use the cosine rule since we have the triangle \( ADC \) with sides \( AD = 4 \, \text{cm} \), \( AC = 6 \, \text{cm} \), and we need to find the angle between them.

Using the cosine rule:
\[
\cos(\angle DAC) = \frac{AD^2 + AC^2 - DC^2}{2 \cdot AD \cdot AC}
\]
Since \( DC \) is also equal to 6 cm (as \( DB = DC \)), I'll now calculate \( \angle DAC \).### (b) The size of \( \angle DAC \) is \( 1.231 \, \text{radians} \), rounded to 3 significant figures.

Next, let's calculate the complete area of the logo design for part (c). The total area includes:
1. The area of the sector \( BAC \) (which we already calculated as \( 39.6 \, \text{cm}^2 \)).
2. The areas of the two triangles \( ABD \) and \( ACD \).

I'll now compute the areas of triangles \( ABD \) and \( ACD \) using the formula:
\[
\text{Area of triangle} = \frac{1}{2} \cdot a \cdot b \cdot \sin(\theta)
\]
where \( a \) and \( b \) are the sides, and \( \theta \) is the angle between them.### (c) The complete area of the logo design is \( 62 \, \text{cm}^2 \), rounded to the nearest cm².

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To summarize:
1. The area of the sector \( BAC \) is \( 39.6 \, \text{cm}^2 \).
2. The size of \( \angle DAC \) is \( 1.231 \, \text{radians} \).
3. The total area of the logo design is \( 62 \, \text{cm}^2 \).

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Do you need further details on any part? Here are 5 related questions to expand your understanding:

1. How is the area of a sector related to the angle and radius?
2. How can the cosine rule be used to solve triangles in general?
3. What would happen if the angle \( \angle BAC \) were smaller? How would it affect the sector's area?
4. How would you calculate the perimeter of the logo design?
5. How does trigonometry help in calculating areas of non-right triangles?

Tip: Always remember that trigonometric identities like the sine and cosine rules are particularly useful in non-right triangles where basic formulas might not apply.

The shape BCD shown above is a design for a logo. The straight lines DB and DC are equal in length. The curve BC is an arc of a circle with center A and radius 6 cm. The size of ∠BAC is 2.2 radians and AD = 4 cm. Find: (a) the area of the sector BAC, in cm²; (b) the size of ∠DAC, in radians to 3 significant figures; (c) the complete area of the logo design, to the nearest cm².

Learn how to calculate the area of a sector, the size of an angle using the cosine rule, and the complete area of a logo design. This problem covers advanced geometry and trigonometry concepts, ideal for students in Grades 10-12.

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