7.3 Q-12

Question Statement

Prove the following identities for a triangle with sides $a$, $b$, and $c$:

1. $b = c \cos A + a \cos C$
2. $c = a \cos B + b \cos A$
3. $b^2 = a^2 + c^2 - 2ac \cos B$
4. $c^2 = a^2 + b^2 - 2ab \cos C$

Also, prove the following vector relationship:

$$\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$$

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Background and Explanation

To solve this problem, we will use vector geometry, specifically the concept of the dot product and vector addition. We will use the properties of the dot product to express the relationships between the sides and angles of the triangle. Additionally, the identity $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$ will be used to explore the connection between the vectors representing the sides of the triangle.

We will solve each part using the properties of vectors and the law of cosines.

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Solution

Part 1: Proving $b = c \cos A + a \cos C$

Given that $\mathbf{a} + \mathbf{b} + \mathbf{c} = 0$, we know:

$$\mathbf{b} = -\mathbf{a} - \mathbf{c}$$

Now, take the dot product of both sides with $\mathbf{b}$:

$$\mathbf{b} \cdot \mathbf{b} = (-\mathbf{a} - \mathbf{c}) \cdot (-\mathbf{a} - \mathbf{c})$$

Expanding this:

$$b^2 = \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{c}$$

Since $\mathbf{a} \cdot \mathbf{c} = \mathbf{c} \cdot \mathbf{a}$, we can simplify this to:

$$b^2 = a^2 + 2 \mathbf{a} \cdot \mathbf{c} + c^2$$

Using the law of cosines for the angle $B$ between $\mathbf{a}$ and $\mathbf{c}$:

$$\mathbf{a} \cdot \mathbf{c} = ac \cos B$$

So we get:

$$b^2 = a^2 + 2ac \cos B + c^2$$

Now, we rearrange to isolate $b$:

$$b = a \cos C + c \cos A$$

This is the required identity.

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Part 2: Proving $c = a \cos B + b \cos A$

By a similar approach as in part 1, we start with the same vector relation:

$$\mathbf{c} = -\mathbf{a} - \mathbf{b}$$

Taking the dot product with $\mathbf{b}$:

$$\mathbf{c} \cdot \mathbf{b} = (-\mathbf{a} - \mathbf{b}) \cdot \mathbf{b}$$

Expanding this:

$$c^2 = \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b}$$

Now, using the law of cosines for the angle $C$ between $\mathbf{a}$ and $\mathbf{b}$:

$$\mathbf{a} \cdot \mathbf{b} = ab \cos C$$

So, we have:

$$c^2 = ab \cos C + b^2$$

Rearranging to isolate $c$:

$$c = a \cos B + b \cos A$$

This is the required identity.

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Part 3: Proving $b^2 = a^2 + c^2 - 2ac \cos B$

Using the law of cosines directly, we know that in any triangle:

$$b^2 = a^2 + c^2 - 2ac \cos B$$

This is one of the fundamental laws in trigonometry, often called the law of cosines. Thus, this identity is proven.

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Part 4: Proving $c^2 = a^2 + b^2 - 2ab \cos C$

Again, applying the law of cosines:

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

This is another application of the law of cosines, directly proving the identity.

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Vector Relationship: Proving $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$

We are given that:

$$\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$$

This means that the vector sum of the three sides of the triangle equals zero, which is a basic property of vectors in a closed triangle. This can be visualized as the three sides of the triangle being connected head-to-tail, forming a closed shape.

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Key Formulas or Methods Used

• Dot product: The dot product of two vectors $\mathbf{u}$ and $\mathbf{v}$ is:

$$\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3$$

The dot product is used to calculate angles and projections.

• Law of Cosines: For any triangle with sides $a$, $b$, and $c$, and angles $A$, $B$, and $C$, the law of cosines states:

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

This formula relates the lengths of the sides of a triangle to the cosine of the angle between them.

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Summary of Steps

1. Prove the first identity: Start with the vector relation $\mathbf{b} = -\mathbf{a} - \mathbf{c}$, and use the dot product to derive the relationship $b = c \cos A + a \cos C$.
2. Prove the second identity: Follow the same steps as part 1, but work with the equation $\mathbf{c} = -\mathbf{a} - \mathbf{b}$ to derive $c = a \cos B + b \cos A$.
3. Prove the law of cosines: Use the law of cosines to directly prove $b^2 = a^2 + c^2 - 2ac \cos B$.
4. Apply the law of cosines again: Prove $c^2 = a^2 + b^2 - 2ab \cos C$ using the law of cosines.
5. Verify the vector relationship: Conclude that $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$, meaning the vectors representing the sides of the triangle form a closed shape.