Law of Cosines Proof

1. Diagram
   Observation

   Triangle side labeled a on the left slanted side.

2. Audio
   Observation

   Speaker says the goal is to solve for side length a.

a

Side length opposite angle A; the unknown side to be solved for.

Positive real number (side length).

1. Diagram
   Observation

   Triangle side labeled b on the right slanted side.

2. Audio
   Observation

   Speaker says we are given side length b.

b

Given side length adjacent to angle A on the right side of the triangle.

Positive real number (side length).

1. Diagram
   Observation

   Base of the triangle labeled c.

2. Audio
   Observation

   Speaker says we are given side length c and later says the whole bottom part is c.

c

Given base side length of the non-right triangle.

Positive real number (side length).

1. Diagram
   Observation

   Angle at the lower-right vertex marked A.

2. Audio
   Observation

   Speaker says we are given angle A.

A

Given interior angle at the lower-right vertex of the triangle.

Interior angle of a triangle.

1. Diagram
   Observation

   Vertical altitude segment labeled h.

2. Audio
   Observation

   Speaker says, let's go ahead and call this altitude h for height.

h

Length of the altitude dropped from the top vertex perpendicular to the base c.

Positive real number (height).

1. Diagram
   Observation

   Right portion of the base labeled x.

2. Audio
   Observation

   Speaker says we're going to go ahead and call this side length x.

x

Length of the right subsegment of the base cut off by the altitude foot.

Positive real number (segment length).

1. Diagram
   Observation

   Left portion of the base labeled c - x.

2. Audio
   Observation

   Speaker says this side length over here has to be c minus x.

c-x

Length of the left subsegment of the base after splitting c into x and c-x.

Positive real number (segment length).

1. Diagram
   Observation

   The left slanted side of the large triangle is labeled a in blue.

2. Audio
   Observation

   The speaker says he is ultimately looking for an expression for a.

a

Side length opposite angle A; the target unknown to solve for.

positive real length

1. Diagram
   Observation

   The right slanted side of the large triangle is labeled b in yellow.

2. Formula
   Observation

   Appears in $b^2 = h^2 + x^2$ and $b^2 - h^2 = x^2$.

b

Side length adjacent to angle A on the right sub-triangle.

positive real length

1. Diagram
   Observation

   The full base of the large triangle is labeled c.

2. Formula
   Observation

   Appears in (c-x)^2 and $c^2 - 2$cx + $x^2$.

c

Full base length of the large triangle, split into segments c-x and x by the altitude foot.

positive real length

1. Diagram
   Observation

   The lower-right vertex is labeled A.

2. Caption evidence
   Observation

   Top text reads Given C, b, and A.

1. The top handwritten text uses uppercase C where the diagram uses lowercase c for the base; the spoken derivation consistently treats c as the base length.

A

Angle at the lower-right vertex of the large triangle.

triangle angle

1. Diagram
   Observation

   The vertical altitude from the top vertex to the base is labeled h.

2. Formula
   Observation

   Appears in $a^2 = (c-x)^2 + h^2$ and $b^2 = h^2 + x^2$.

h

Height of the perpendicular dropped from the top vertex to the base.

positive real length

1. Audio
   Observation

   Speaker states that the law of cosines is important because it makes solving non-right triangles easier and motivates proving why the equation works.

2. Diagram
   Observation

   Opening title reads Law of Cosines Proof.

The video introduces the law of cosines as a tool for solving non-right triangles and frames the clip as a proof of where the formula comes from.

1. Applies to non-right triangles.

2. Used when two sides and the included angle are known in this setup.

1. Diagram
   Observation

   Text on screen: Given c, b, and A; find an expression to solve for a.

2. Audio
   Observation

   Speaker says we are given side length c, side length b, and angle A, and the ultimate goal is to solve for side length a.

The proof starts from a triangle with known sides c and b and known included angle A, and seeks an expression for the remaining side a.

1. Known quantities are c, b, and A.

2. Target unknown is a.

1. Purpose of the law of cosines

1. Audio
   Observation

   Speaker says dealing with non-right triangles is difficult, so the first step is to drop an altitude from the very top to make a perpendicular line to the base.

2. Diagram
   Observation

   A vertical segment is drawn from the top vertex to the base, with a right-angle mark at the foot.

The method converts one non-right triangle into two right triangles by drawing a perpendicular altitude from the top vertex to the base.

1. Start with a non-right triangle.

2. Drop a perpendicular from the top vertex to the base.

1. Given information and goal of the proof

1. Audio
   Observation

   Speaker says now we've created two right triangles from one non-right triangle and outlines them in bold lines.

2. Diagram
   Observation

   Left triangle outlined in blue and right triangle outlined in yellow.

After the altitude is drawn, the original triangle is decomposed into a left right triangle and a right right triangle, visually distinguished by blue and yellow outlines.

1. The altitude must meet the base perpendicularly.

2. The decomposition is used to apply right-triangle tools.

1. Dropping an altitude to create right triangles

1. Audio
   Observation

   Speaker names the altitude h, calls the right base piece x, and says the other piece must be c minus x because the whole length is c.

2. Diagram
   Observation

   Labels h, x, and c-x appear on the altitude and the two base segments.

The altitude is named h. The base c is split into a right segment x and a left segment c-x, so that the two new right triangles have legs expressible in terms of h, x, and c-x.

1. Whole base length is c.

2. Right subsegment is x.

3. Left subsegment is c-x.

1. Two right triangles from one altitude

1. Audio
   Observation

   Speaker says because we have two right triangles, we're allowed to use Pythagorean theorem and trig functions, SOHCAHTOA, as long as we use them with separate right triangles.

2. Diagram
   Observation

   Blue and yellow right triangles remain visible with labeled legs and hypotenuses.

Once the figure is decomposed into right triangles, the proof can use the Pythagorean theorem and right-triangle trigonometric ratios on each smaller triangle separately.

1. Must work within each right triangle separately.

2. The altitude creates the required right angles.

1. Naming the altitude and split base segments

1. Audio
   Observation

   I know the hypotenuse squared is equal to the two legs of the triangle squared, c minus x squared added together, h squared.

2. Formula
   Observation

   $a^2 = (c-x)^2 + h^2$ is written under the blue left sub-triangle.

Applying the Pythagorean theorem to the blue right triangle gives the square of its hypotenuse a as the sum of the squares of its legs c-x and h.

1. The altitude from the top vertex meets the base at a right angle.

2. a is the hypotenuse of the left sub-triangle.

1. Pythagorean theorem used as analogy

1. Audio
   Observation

   This is essentially a squared plus b squared equals c squared, just looks a little bit more fancy.

The speaker identifies the previous equation as an instance of the standard right-triangle identity, with the roles of the sides renamed by the geometry of the construction.

1. Applies to right triangles.

1. Audio
   Observation

   Let's go ahead and expand this. I'm going to multiply c minus x times c minus x.

2. Formula
   Observation

   (c-x)(c-x) = $c^2$ - cx - cx + $x^2 = c^2 - 2$cx + $x^2$ is written on the right.

The square of the binomial is expanded by distributing each term, producing two middle terms that combine into -2cx.

1. Standard algebra over real numbers.

1. Audio
   Observation

   And we still have that plus h squared off to the side.

2. Formula
   Observation

   $a^2 = c^2 - 2$cx + $x^2 + h^2$ is written below the first equation.

Replacing (c-x)^2 in the Pythagorean equation yields an expanded expression for $a^2$ that still contains the auxiliary variables x and h.

1. Follows directly from $a^2 = (c-x)^2 + h^2$ and (c-x)^$2 = c^2 - 2$cx + $x^2$.

1. Pythagorean relation for the left sub-triangle
2. Expansion of (c-x)^2 by FOIL

1. Audio
   Observation

   So I'm going to start by looking at this yellow triangle over here now. And I'm going to go ahead and do Pythagorean theorem for that as well.

2. Formula
   Observation

   $b^2 = h^2 + x^2$ is written under the yellow right sub-triangle.

Applying the same theorem to the yellow right triangle relates its hypotenuse b to the altitude h and base segment x.

1. The altitude from the top vertex meets the base at a right angle.

2. b is the hypotenuse of the right sub-triangle.

1. Pythagorean theorem used as analogy

1. Audio
   Observation

   If I solve for x squared, then I see that x squared is equal to this b squared minus h squared.

2. Formula
   Observation

   $b^2 - h^2 = x^2$ is written beneath $b^2 = h^2 + x^2$.

Rearranging $b^2 = h^2 + x^2$ expresses $x^2$ in terms of b and h, preparing for elimination of auxiliary variables from the equation for $a^2$.

1. Equivalent algebraic rearrangement of $b^2 = h^2 + x^2$.

1. Pythagorean relation for the right sub-triangle

1. Audio
   Observation

   Speaker says the law of cosines is extremely important because it makes solving non-right triangles a lot easier.

The law of cosines is useful because it makes solving non-right triangles easier.

1. Working with non-right triangles.

General statement about the utility of the law of cosines.

1. Audio
   Observation

   Speaker says dropping an altitude from the top perpendicular to the base creates two right triangles from one non-right triangle.

2. Diagram
   Observation

   A perpendicular segment is drawn from the top vertex to the base, with a right-angle marker.

Dropping an altitude from the top vertex perpendicular to the base of a non-right triangle creates two right triangles.

1. Start with a non-right triangle.

2. Draw a perpendicular from the top vertex to the base.

For the triangle shown in the proof setup.

1. Audio
   Observation

   Speaker says if the whole length is c and this part is x, then the other part has to be c minus x.

2. Diagram
   Observation

   Base labeled c overall, with subsegments labeled x and c-x.

If the full base length is c and one subsegment is x, then the other subsegment is c-x.

1. The altitude foot splits the base into two pieces.

2. The whole base length is c.

3. One piece is named x.

For the two base subsegments in the diagram.

1. Audio
   Observation

   Speaker says because we have two right triangles, we're allowed to use Pythagorean theorem and trig functions, SOHCAHTOA, as long as we use them with separate right triangles.

After the altitude is drawn, the Pythagorean theorem and right-triangle trigonometric ratios may be applied to each of the two smaller right triangles separately.

1. The original triangle has been split into two right triangles by an altitude.

For each of the two right triangles created in the construction.

1. Audio
   Observation

   Part of the reason that I know to do that is because I'm ultimately looking for an expression for a. Well now I have a squared equals something over here. So if I can find a way to get rid of these x's and h's, then I have an equation for a.

Once $a^2$ has been rewritten in terms of c, x, and h, eliminating the auxiliary variables x and h will yield an equation involving only the desired side a and the given quantities.

1. The goal is to solve for a.

2. The current equation is $a^2 = c^2 - 2$cx + $x^2 + h^2$.

For the triangle setup shown, if x and h can be removed from the equation, then the remaining relation determines a.

1. Audio
   Observation

   The speaker explicitly says he went from having three unknowns in his equation to just one.

2. Formula
   Observation

   The displayed progression moves from $a^2 = c^2 - 2$cx + $x^2 + h^2$ to $a^2 = c^2 - 2$cx + $b^2$.

Using $x^2 = b^2 - h^2$ to replace $x^2$ in the expanded left-triangle equation eliminates h and leaves only x as the remaining auxiliary unknown.

1. The two right-triangle Pythagorean relations are valid.

2. Algebraic substitution and cancellation are allowed.

For the triangle configuration shown on the board.

1. Audio
   Observation

   The speaker calls the result the derivation or proof of the law of cosines.

2. Formula
   Observation

   The final boxed equation is $a^2 = b^2 + c^2 - 2$bc $\cos (A)$.

1. The visible proof is written for the drawn acute configuration with the altitude falling inside the base; the clip does not separately discuss obtuse or right cases.

For the triangle shown, the side opposite angle A satisfies $a^2 = b^2 + c^2 - 2$bc\cos(A).

1. A triangle has side lengths a, b, c and angle A as labeled.

2. An altitude h is dropped to the base, splitting the base into segments x and c-x.

For the labeled triangle configuration shown in the clip.

1. Audio
   Observation

   Speaker explains the strategy: drop an altitude, create two right triangles, name the altitude h, and split the base into x and c-x.

2. Diagram
   Observation

   Construction appears step by step: altitude, right-angle mark, blue/yellow outlines, labels h, x, c-x.

1. Expression

   $$$$
   Explanation

   Start with a non-right triangle whose known parts are sides c and b and included angle A, with unknown side a.

   Justification

   This is the stated given information and goal of the proof.

   Shown in the video
2. Expression

   $$$$
   Explanation

   Drop a perpendicular altitude from the top vertex to the base c.

   Justification

   The speaker says this is the usual first step for handling non-right triangles.

   Shown in the video
3. Expression

   $$$$
   Explanation

   The altitude creates two right triangles inside the original triangle.

   Justification

   A perpendicular to the base produces right angles at the foot of the altitude.

   Shown in the video
4. Expression

   $$h$$
   Explanation

   Name the altitude h.

   Justification

   The speaker explicitly assigns this label for convenience.

   Shown in the video
5. Expression

   $$x$$
   Explanation

   Name the right-hand base subsegment x.

   Justification

   The speaker introduces x as a semi-arbitrary label for the proof.

   Shown in the video
6. Expression

   $$c-x$$
   Explanation

   Then the left-hand base subsegment is c-x.

   Justification

   Since the whole base is c and one part is x, the remaining part must be c-x.

   Shown in the video

The original non-right triangle has been rewritten as two right triangles with legs h and x on the right and h and c-x on the left, setting up the use of Pythagorean theorem and trigonometry.

1. Audio
   Observation

   Speaker says because we have two right triangles, we're allowed to use Pythagorean theorem and trig functions, SOHCAHTOA, as long as we use them with separate right triangles.

2. Diagram
   Observation

   Blue triangle on the left and yellow triangle on the right remain highlighted.

1. Expression

   $$$$
   Explanation

   Recognize that the construction has produced two right triangles rather than one oblique triangle.

   Justification

   The altitude is perpendicular to the base, so each smaller triangle contains a right angle.

   Derived from the video
2. Expression

   $$$$
   Explanation

   Therefore the Pythagorean theorem is now applicable to each smaller triangle separately.

   Justification

   Pythagorean theorem applies to right triangles.

   Shown in the video
3. Expression

   $$$$
   Explanation

   Also, right-triangle trigonometric ratios such as SOHCAHTOA are applicable within each smaller triangle.

   Justification

   The speaker explicitly states that trig functions are fair game once separate right triangles are available.

   Shown in the video

The proof is ready to proceed by applying right-triangle relations to the blue and yellow triangles individually.

1. Formula
   Observation

   The board shows $a^2 = (c-x)^2 + h^2$, then (c-x)(c-x), then $c^2$ - cx - cx + $x^2$, then $c^2 - 2$cx + $x^2$, then $a^2 = c^2 - 2$cx + $x^2 + h^2$.

2. Audio
   Observation

   The speaker narrates applying the Pythagorean theorem, expanding with distributive property or FOIL, and keeping + $h^2$ off to the side.

1. Expression

   $$a^2 = (c-x)^2 + h^2$$
   Explanation

   Apply the Pythagorean theorem to the blue right triangle whose legs are c-x and h and whose hypotenuse is a.

   Justification

   Pythagorean theorem on a right triangle.

   Shown in the video
2. Expression

   $$(c-x)^2 = (c-x)(c-x)$$
   Explanation

   Rewrite the squared binomial as a product so it can be distributed.

   Justification

   Definition of squaring a binomial.

   Shown in the video
3. Expression

   $$(c-x)(c-x) = c^2 - cx - cx + x^2$$
   Explanation

   Multiply term by term using FOIL / distributive property.

   Justification

   Distributive property of multiplication over addition/subtraction.

   Shown in the video
4. Expression

   $$c^2 - cx - cx + x^2 = c^2 - 2cx + x^2$$
   Explanation

   Combine the two identical middle terms.

   Justification

   Combining like terms.

   Shown in the video
5. Expression

   $$a^2 = c^2 - 2cx + x^2 + h^2$$
   Explanation

   Substitute the expanded form back into the original Pythagorean equation.

   Justification

   Substitution of equals for equals.

   Shown in the video

The left-side relation becomes $a^2 = c^2 - 2$cx + $x^2 + h^2$.

1. Formula
   Observation

   The board shows $b^2 = h^2 + x^2$ and then $b^2 - h^2 = x^2$.

2. Audio
   Observation

   The speaker says he will do Pythagorean theorem for the yellow triangle and then solve for x squared.

1. Expression

   $$b^2 = h^2 + x^2$$
   Explanation

   Apply the Pythagorean theorem to the yellow right triangle with legs h and x and hypotenuse b.

   Justification

   Pythagorean theorem on a right triangle.

   Shown in the video
2. Expression

   $$b^2 - h^2 = x^2$$
   Explanation

   Move $h^2$ to the other side to isolate $x^2$.

   Justification

   Subtract $h^2$ from both sides of the equation.

   Shown in the video

The auxiliary quantity $x^2$ is expressed as $b^2 - h^2$.

1. Audio
   Observation

   Continuous spoken derivation from substitution of $x^2$ through cosine elimination of x.

2. Formula
   Observation

   Sequence of displayed equations: $a^2=(c-x)^2+h^2$, $b^2=h^2+x^2$, $b^2-h^2=x^2$, $a^2=c^2-2$cx+$x^2+h^2$, $a^2=c^2-2$cx+$b^2-h^2+h^2$, $a^2=c^2-2$cx+$b^2$, $\cos A=x/b$, $b \cos (A)=x$, $a^2=c^2-2$cb $\cos (A)+b^2$.

1. The clip begins after some setup has already been written, so the initial construction of the altitude is not fully shown within this excerpt.

1. Expression

   $$a^2 = (c-x)^2 + h^2$$
   Explanation

   Start from the left right triangle formed by the altitude.

   Justification

   Pythagorean theorem applied to the triangle with hypotenuse a and legs c-x and h.

   Shown in the video
2. Expression

   $$b^2 = h^2 + x^2$$
   Explanation

   Write the corresponding relation for the right right triangle.

   Justification

   Pythagorean theorem applied to the triangle with hypotenuse b and legs h and x.

   Shown in the video
3. Expression

   $$b^2 - h^2 = x^2$$
   Explanation

   Rearrange the second relation to isolate $x^2$.

   Justification

   Algebraic subtraction of $h^2$ from both sides.

   Shown in the video
4. Expression

   $$a^2 = c^2 - 2cx + x^2 + h^2$$
   Explanation

   Expand the square in the first relation.

   Justification

   Binomial expansion of (c-x)^2.

   Shown in the video
5. Expression

   $$a^2 = c^2 - 2cx + b^2 - h^2 + h^2$$
   Explanation

   Replace $x^2$ by $b^2-h^2$ in the expanded equation.

   Justification

   Substitution using the isolated expression for $x^2$.

   Shown in the video
6. Expression

   $$a^2 = c^2 - 2cx + b^2$$
   Explanation

   Cancel the opposite $h^2$ terms.

   Justification

   Additive inverse cancellation: -h^$2 + h^2 = 0$.

   Shown in the video
7. Expression

   $$\cos A = \frac{x}{b}$$
   Explanation

   Express x using trigonometry in the right triangle containing angle A.

   Justification

   Definition of cosine as adjacent over hypotenuse.

   Shown in the video
8. Expression

   $$b\cos(A) = x$$
   Explanation

   Solve the cosine equation for x.

   Justification

   Multiply both sides by b.

   Shown in the video
9. Expression

   $$a^2 = c^2 - 2cb\cos(A) + b^2$$
   Explanation

   Substitute $x = b \cos (A)$ into the reduced equation.

   Justification

   Direct substitution into the term -2cx.

   Shown in the video

The displayed derivation obtains the Law of Cosines in the form $a^2 = c^2 - 2$cb\cos(A) + $b^2$ for the triangle shown.

1. Formula
   Observation

   The board displays the sequence $a^2=(c-x)^2+h^2$, then expansion, substitution of $b^2-h^2$ for $x^2$, cancellation of $h^2$, simplification to $a^2=c^2-2$cx+$b^2$, and substitution $x=b \cos (A)$.

2. Audio
   Observation

   The speaker narrates the rearrangement into standard form and summarizes the proof steps.

1. The earlier algebraic lines are already present at the start of the clip, so the exact moment when each prior line was first written is not visible in this excerpt.

1. Expression

   $$a^2 = (c-x)^2 + h^2$$
   Explanation

   Apply the Pythagorean theorem to the left right triangle, whose legs are h and c-x and whose hypotenuse is a.

   Justification

   Pythagorean theorem on the left triangle.

   Shown in the video
2. Expression

   $$a^2 = c^2 - 2cx + x^2 + h^2$$
   Explanation

   Expand (c-x)^2 and keep the $h^2$ term.

   Justification

   Algebraic expansion of a square.

   Shown in the video
3. Expression

   $$a^2 = c^2 - 2cx + (b^2 - h^2) + h^2$$
   Explanation

   Replace $x^2$ using the right-triangle relation $b^2 = h^2 + x^2$, which gives $x^2 = b^2 - h^2$.

   Justification

   Substitution from the Pythagorean theorem on the right triangle.

   Shown in the video
4. Expression

   $$a^2 = c^2 - 2cx + b^2$$
   Explanation

   Cancel +$h^2$ and -h^2.

   Justification

   Algebraic simplification.

   Shown in the video
5. Expression

   $$x = b\cos A$$
   Explanation

   From the right triangle containing angle A, use the cosine ratio $\cos A = x/b$.

   Justification

   Definition of cosine in a right triangle (SOHCAHTOA).

   Shown in the video
6. Expression

   $$a^2 = c^2 - 2cb\cos(A) + b^2$$
   Explanation

   Substitute $x = b \cos A$ into the previous line.

   Justification

   Direct substitution.

   Shown in the video
7. Expression

   $$a^2 = b^2 + c^2 - 2bc\cos(A)$$
   Explanation

   Reorder terms into the standard law-of-cosines form.

   Justification

   Commutative rearrangement of addition.

   Shown in the video

The clip derives the law of cosines in the form $a^2 = b^2 + c^2 - 2$bc\cos(A).

1. Diagram
   Observation

   Black background with handwritten white title Law of Cosines Proof.

1. Handwritten title text

1. Static title remains on screen while the speaker introduces the topic.

1. No triangle diagram yet.

The clip is framed as a proof of the law of cosines rather than just formula usage.

1. Diagram
   Observation

   A scalene triangle appears with labels a, b, c, and angle A; text reads Given c, b, and A and find an expression to solve for a.

1. Triangle

2. Labels a, b, c, A

3. Prompt text

1. The triangle and givens are introduced all at once.

1. The target unknown is side a.

2. The known data are c, b, and A.

This establishes the SAS-style configuration used for deriving the law of cosines.

1. Diagram
   Observation

   A vertical segment is drawn from the top vertex to the base, followed by a small square right-angle marker at the foot.

1. Original triangle

2. Altitude segment

3. Right-angle marker

1. A perpendicular segment is added from the top vertex to the base.

2. A right-angle symbol marks the foot of the altitude.

1. The outer triangle sides a, b, c and angle A remain the same.

The construction converts the oblique triangle problem into a pair of right-triangle problems.

1. Diagram
   Observation

   The left inner triangle is outlined in blue and the right inner triangle in yellow.

1. Blue-outlined left right triangle

2. Yellow-outlined right right triangle

1. Color emphasis separates the two newly formed right triangles.

1. Both share the same altitude as a common vertical side.

The color coding clarifies which parts belong to which right triangle for the upcoming algebra.

1. Diagram
   Observation

   Labels h, x, and c-x are written on the altitude and the two base segments.

1. Altitude label h

2. Right base label x

3. Left base label c-x

1. New symbolic labels are added to the geometric construction.

1. The total base remains $c = x + (c-x)$.

The diagram now expresses both right triangles entirely in terms of h, x, c-x, a, and b.

1. Diagram
   Observation

   A large triangle is drawn with base c, left side a, right side b, top vertex connected to the base by altitude h, and the base split into c-x on the left and x on the right; a right-angle mark appears at the foot of the altitude.

2. Caption evidence
   Observation

   Top text reads Given C, b, and A and find an expression to solve for a.

1. The top text writes C while the diagram labels the base as c.

1. large triangle

2. left sub-triangle

3. right sub-triangle

4. altitude h

5. base segments c-x and x

6. right-angle marker

7. labels a, b, c, A, h, x

1. Equations are written sequentially beneath and beside the figure as the derivation proceeds.

1. The altitude remains perpendicular to the base throughout.

2. The decomposition $c = (c-x) + x$ stays fixed in the diagram.

The picture encodes two right triangles sharing the same altitude h, which allows the law-of-cosines derivation to proceed by applying the Pythagorean theorem separately to each sub-triangle.

1. Diagram
   Observation

   Curved arrows connect the terms in (c-x)(c-x) to show first, outer, inner, and last products.

1. (c-x)(c-x)

2. curved multiplication arrows

1. Arrows indicate pairwise multiplication of terms before the expanded polynomial is written.

1. The product being expanded remains (c-x)(c-x).

The arrow diagram makes explicit that the middle terms arise from the outer and inner products, both equal to -cx.

1. Diagram
   Observation

   The cursor points near the terms involving x and h in the equation $a^2 = c^2 - 2$cx + $x^2 + h^2$ while the speaker discusses getting rid of these x's and h's.

1. equation $a^2 = c^2 - 2$cx + $x^2 + h^2$

2. cursor

1. Attention shifts to the x and h terms as the next algebraic target.

1. The equation itself does not change during this moment.

This visual cue marks the transition from deriving an expanded formula to planning variable elimination.

1. Diagram
   Observation

   A large triangle is drawn with a vertical altitude splitting it into a blue left right triangle and a yellow right right triangle.

2. Diagram
   Observation

   Labels a, b, c, h, x, c-x, and A remain visible throughout the clip.

1. large triangle

2. altitude h

3. left right triangle

4. right right triangle

5. labels a,b,c,h,x,c-x,A

1. New algebra lines are added below the diagram as the proof proceeds.

2. The cosine relation is added on the right side of the board late in the clip.

1. The geometric decomposition into two right triangles stays fixed.

2. The labels on the triangle do not change during the excerpt.

The visual layout encodes the proof strategy: use the altitude to create two right triangles, relate their sides by the Pythagorean theorem, then eliminate auxiliary lengths using the given angle A.

1. Audio
   Observation

   The speaker says he will use color to help viewers see the substitution and later says 'wrong color, green'.

2. Diagram
   Observation

   The substituted terms are written in contrasting colors on the new lines.

1. Exact color assignment of every symbol is not fully stable across frames, but the speaker clearly uses color to mark substituted pieces.

1. $x^2$ term

2. $b^2-h^2$ substitution

3. $b \cos (A)$ substitution

1. The $x^2$ term is visually replaced by $b^2-h^2$.

2. Later the x in -2cx is replaced by $b \cos (A)$.

1. The underlying algebraic equality is unchanged except for the indicated substitutions.

Color is used pedagogically to track which subexpression is being replaced at each step of the derivation.

1. Diagram
   Observation

   A black digital whiteboard shows a large triangle at upper left, a vertical altitude h, base segments c-x and x, and multiple handwritten equations below and to the right.

1. Large triangle labeled a, b, c

2. Altitude h

3. Base segments c-x and x

4. Angle A at lower right

5. Handwritten equations below the triangle

6. Green cosine relations at upper right

7. Blue boxed final formula at lower right

1. The main new writing during the clip is the blue final formula $a^2 = b^2 + c^2 - 2$bc $\cos (A)$.

2. A blue box is drawn around the final formula near the end.

1. The triangle diagram and earlier derivation lines remain visible throughout the clip.

The visual organization separates geometric setup, intermediate algebra, trigonometric substitution, and the final standard-form result.

1. Animation
   Observation

   After the final formula is completed, a blue rectangular outline is drawn around it.

1. Final formula $a^2 = b^2 + c^2 - 2$bc $\cos (A)$

2. Blue rectangular box

1. The completed formula becomes enclosed in a box.

1. The formula content itself does not change after boxing.

The boxing marks the formula as the concluding result of the proof.

1. Audio
   Observation

   Speaker says dealing with non-right triangles is pretty difficult, so usually the first step is to take this non-right triangle and drop an altitude.

One might expect to apply right-triangle formulas directly to the original non-right triangle.

The video instead first converts the figure into two right triangles by dropping an altitude.

1. Audio
   Observation

   Speaker says we're allowed to use Pythagorean theorem and those trig functions, so SOHCAHTOA, both of those are fair game now as long as we use them with separate right triangles.

After introducing trig ratios and Pythagorean theorem, one might mix relations across the whole original triangle.

The speaker explicitly restricts their use to the separate right triangles created by the altitude.

1. Audio
   Observation

   This is essentially a squared plus b squared equals c squared, just looks a little bit more fancy.

Because the equation is written as $a^2 = (c-x)^2 + h^2$, one may think it is no longer the Pythagorean theorem.

It is exactly the Pythagorean theorem applied to a right triangle whose legs are c-x and h and whose hypotenuse is a.

1. Audio
   Observation

   I know that this is very counterintuitive, but let's go ahead and expand this.

One might try to manipulate $a^2 = (c-x)^2 + h^2$ without expanding (c-x)^2 first.

The video explicitly expands (c-x)^2 into $c^2 - 2$cx + $x^2$ so the later cancellation and substitution can be seen.

1. Audio
   Observation

   The speaker contrasts having x, $x^2$, and $h^2$ with ending at only one remaining unknown.

One might think the equation still has several independent unknowns after expansion.

The video emphasizes that substitution removes $h^2$ entirely and later removes x, leaving an expression for $a^2$ in terms of the given quantities b, c, and A.

1. Diagram
   Observation

   The drawn altitude falls inside the base, matching an acute-angle picture.

2. Audio
   Observation

   The speaker summarizes the proof without discussing other configurations.

One might think the displayed argument explicitly treats all possible triangle shapes in this clip.

The clip visibly proves the formula for the drawn configuration with an internal altitude; it does not separately discuss obtuse or right-triangle cases in this excerpt.

1. Audio
   Observation

   Speaker moves from saying the law of cosines helps with non-right triangles to stating the specific givens c, b, A and target a.

The general usefulness of the law of cosines is instantiated by the specific triangle-solving goal shown on screen.

1. Audio
   Observation

   After stating the givens, the speaker says the usual first step for non-right triangles is to drop an altitude.

The given oblique-triangle problem motivates the altitude construction.

1. Audio
   Observation

   Speaker says dropping the altitude creates two right triangles from one non-right triangle.

2. Diagram
   Observation

   Blue and yellow outlines show the two resulting right triangles.

The altitude method produces the pair of right triangles as its immediate geometric result.

1. Audio
   Observation

   Speaker names h, x, and c-x after the two right triangles have been identified.

2. Diagram
   Observation

   Labels are added to the altitude and base segments.

Describing the two right triangles requires naming the shared altitude and the split base segments.

1. Audio
   Observation

   Speaker says because we have two right triangles, we're allowed to use Pythagorean theorem and trig functions.

Only after the figure is expressed as two right triangles with labeled legs can Pythagorean theorem and SOHCAHTOA be applied in the proof.

1. Audio
   Observation

   The altitude is introduced specifically so the non-right triangle can be handled with right-triangle tools.

The availability of Pythagorean theorem in the proof depends on first constructing the altitude.

1. Audio
   Observation

   The speaker says the equation is essentially a squared plus b squared equals c squared.

2. Formula
   Observation

   $a^2 = (c-x)^2 + h^2$ is written from the blue sub-triangle.

The general Pythagorean theorem is instantiated on the left right-triangle formed by the altitude.

1. Audio
   Observation

   I'm going to go ahead and do Pythagorean theorem for that as well.

2. Formula
   Observation

   $b^2 = h^2 + x^2$ is written from the yellow sub-triangle.

The same theorem is applied again to the second right triangle created by the altitude.

1. Formula
   Observation

   The board first derives $c^2 - 2$cx + $x^2$ and then writes $a^2 = c^2 - 2$cx + $x^2 + h^2$.

The substituted equation for $a^2$ depends directly on the prior expansion of (c-x)^2.

1. Audio
   Observation

   If I can find a way to get rid of these x's and h's, then I have an equation for a.

2. Formula
   Observation

   After writing $a^2 = c^2 - 2$cx + $x^2 + h^2$, the speaker derives $b^2 - h^2 = x^2$.

Solving for $x^2$ from the right sub-triangle is presented as the next step toward eliminating auxiliary variables from the equation for $a^2$.

1. Formula
   Observation

   The board moves from $b^2=h^2+x^2$ to $b^2-h^2=x^2$ and then substitutes that into the $a^2$ expansion.

The isolation of $x^2$ depends directly on the right-triangle Pythagorean relation.

1. Formula
   Observation

   After writing $a^2=c^2-2$cx+$b^2-h^2+h^2$, the next line cancels to $a^2=c^2-2$cx+$b^2$.

The cancellation step is only available after substituting $b^2-h^2$ for $x^2$.

1. Audio
   Observation

   Speaker explains the importance of the law of cosines for non-right triangles.

1. Purpose of the law of cosines

1. Diagram
   Observation

   On-screen prompt states Given c, b, and A; find an expression to solve for a.

1. Given information and goal of the proof

1. Audio
   Observation

   Speaker says the usual first step is to drop an altitude from the top of the non-right triangle.

1. Dropping an altitude to create right triangles

1. Diagram
   Observation

   Blue and yellow outlines distinguish the two right triangles.

1. Two right triangles from one altitude

1. Diagram
   Observation

   Labels h, x, and c-x are added to the construction.

1. Naming the altitude and split base segments

1. Audio
   Observation

   Speaker says Pythagorean theorem and SOHCAHTOA are now fair game because there are two right triangles.

1. Why Pythagorean theorem and trig ratios become available

1. Audio
   Observation

   I know the hypotenuse squared is equal to the two legs of the triangle squared.

1. Pythagorean relation for the left sub-triangle
2. Pythagorean theorem used as analogy

1. Formula
   Observation

   (c-x)(c-x) = $c^2 - 2$cx + $x^2$ is shown with FOIL arrows.

1. Expansion of (c-x)^2 by FOIL