Math Problem Statement

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.

Solution

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 aa units long,
  • Side bb units long,
  • Angle CC between the two sides.

We aim to find the third side of the triangle, denoted as cc, using this information.

Step 2: Find the Altitude (Height) Drawn to Side bb

We need to find the length of the altitude (denoted as hh) drawn from the vertex opposite side bb to the base bb. This altitude is perpendicular to side bb.

The altitude hh is related to side aa and angle CC as: h=asin(C)h = a \sin(C) This gives the length of the perpendicular from side aa to side bb.

Step 3: Find the Perpendicular Projection of Side aa onto Side bb

The projection of side aa onto side bb (denoted as pp) is given by: p=acos(C)p = a \cos(C) This is the length of the part of side aa that lies along side bb.

Step 4: Use the Pythagorean Theorem for the Resulting Right Triangle

Now, consider the right triangle formed by:

  • The altitude h=asin(C)h = a \sin(C),
  • The base bp=bacos(C)b - p = b - a \cos(C),
  • The third side of the triangle cc.

We can apply the Pythagorean theorem to this right triangle: c2=h2+(bp)2c^2 = h^2 + (b - p)^2 Substitute the expressions for hh and pp: c2=(asin(C))2+(bacos(C))2c^2 = (a \sin(C))^2 + (b - a \cos(C))^2

Step 5: Simplify the Expression

Now expand and simplify the expression: c2=a2sin2(C)+(bacos(C))2c^2 = a^2 \sin^2(C) + (b - a \cos(C))^2 c2=a2sin2(C)+(b22abcos(C)+a2cos2(C))c^2 = a^2 \sin^2(C) + \left( b^2 - 2ab \cos(C) + a^2 \cos^2(C) \right) c2=a2sin2(C)+b22abcos(C)+a2cos2(C)c^2 = a^2 \sin^2(C) + b^2 - 2ab \cos(C) + a^2 \cos^2(C) Now, recall the trigonometric identity sin2(C)+cos2(C)=1\sin^2(C) + \cos^2(C) = 1: a2sin2(C)+a2cos2(C)=a2a^2 \sin^2(C) + a^2 \cos^2(C) = a^2 Thus, the expression for c2c^2 becomes: c2=a2+b22abcos(C)c^2 = a^2 + b^2 - 2ab \cos(C)

Final Result: Law of Cosines

The third side of the triangle, cc, is given by: c=a2+b22abcos(C)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 CC 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=90C = 90^\circ.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Geometry
Triangles

Formulas

Law of Cosines: c^2 = a^2 + b^2 - 2ab * cos(C)
Trigonometric identity: sin^2(C) + cos^2(C) = 1

Theorems

Law of Cosines

Suitable Grade Level

Grades 10-12