A segment that is a units long makes a C-degree angle with a segment that is b units

long. In terms of a, b, and C, find the third side of the triangle defined by this SAS description.

You have done numerical versions of this question. Start by finding the length of the altitude

drawn to side b, as well as the length of the perpendicular projection of side a onto side b. The

resulting formula is known as the Law of Cosines.

To solve this problem step-by-step, we will derive the third side of the triangle using the Law of Cosines. We'll proceed by first finding the length of the altitude and the perpendicular projection, then use trigonometric relationships to derive the Law of Cosines.

### Step 1: Define the Triangle
We have:
- Side \( a \) units long,
- Side \( b \) units long,
- Angle \( C \) between the two sides.

We aim to find the third side of the triangle, denoted as \( c \), using this information.

### Step 2: Find the Altitude (Height) Drawn to Side \( b \)
We need to find the length of the altitude (denoted as \( h \)) drawn from the vertex opposite side \( b \) to the base \( b \). This altitude is perpendicular to side \( b \).

The altitude \( h \) is related to side \( a \) and angle \( C \) as:
\[
h = a \sin(C)
\]
This gives the length of the perpendicular from side \( a \) to side \( b \).

### Step 3: Find the Perpendicular Projection of Side \( a \) onto Side \( b \)
The projection of side \( a \) onto side \( b \) (denoted as \( p \)) is given by:
\[
p = a \cos(C)
\]
This is the length of the part of side \( a \) that lies along side \( b \).

### Step 4: Use the Pythagorean Theorem for the Resulting Right Triangle
Now, consider the right triangle formed by:
- The altitude \( h = a \sin(C) \),
- The base \( b - p = b - a \cos(C) \),
- The third side of the triangle \( c \).

We can apply the Pythagorean theorem to this right triangle:
\[
c^2 = h^2 + (b - p)^2
\]
Substitute the expressions for \( h \) and \( p \):
\[
c^2 = (a \sin(C))^2 + (b - a \cos(C))^2
\]

### Step 5: Simplify the Expression
Now expand and simplify the expression:
\[
c^2 = a^2 \sin^2(C) + (b - a \cos(C))^2
\]
\[
c^2 = a^2 \sin^2(C) + \left( b^2 - 2ab \cos(C) + a^2 \cos^2(C) \right)
\]
\[
c^2 = a^2 \sin^2(C) + b^2 - 2ab \cos(C) + a^2 \cos^2(C)
\]
Now, recall the trigonometric identity \( \sin^2(C) + \cos^2(C) = 1 \):
\[
a^2 \sin^2(C) + a^2 \cos^2(C) = a^2
\]
Thus, the expression for \( c^2 \) becomes:
\[
c^2 = a^2 + b^2 - 2ab \cos(C)
\]

### Final Result: Law of Cosines
The third side of the triangle, \( c \), is given by:
\[
c = \sqrt{a^2 + b^2 - 2ab \cos(C)}
\]

This is the Law of Cosines formula, which relates the three sides of a triangle and the cosine of the angle between them.

---

Do you have any questions or want more details? Here are 5 related questions you could explore:

1. What happens to the Law of Cosines when the angle \( C \) is a right angle?
2. How is the Law of Cosines used to determine the angles in a triangle when the sides are known?
3. Can you derive the Law of Cosines using vector geometry?
4. What is the relationship between the Law of Cosines and the Pythagorean theorem?
5. How can the Law of Cosines be applied to solve real-world problems involving triangles?

**Tip:** Always remember to check if the triangle is a right triangle, as it simplifies to the Pythagorean theorem when \( C = 90^\circ \).

A segment that is a units long makes a C-degree angle with a segment that is b units long. In terms of a, b, and C, find the third side of the triangle defined by this SAS description.

Learn how to find the third side of a triangle when given two sides and the angle between them using the Law of Cosines. This problem involves trigonometry and geometry concepts, and includes a step-by-step derivation suitable for students in Grades 10-12.

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