Learning math: trigonometry and geometry

These are my solutions to the problems in the book Maths: A Student's Survival Guide.

Right-angled triangles

Remember:

$\sin \alpha = \dfrac{a}{c} \qquad \cos \alpha = \dfrac{b}{c} \qquad \tan \alpha = \dfrac{a}{b}$

If you take the reciprocals of these, you get:

$\dfrac{1}{\sin \alpha} = \dfrac{c}{a} = \cosec \alpha \qquad \dfrac{1}{\cos \alpha} = \dfrac{c}{b} = \sec \alpha \qquad \dfrac{1}{\tan \alpha} = \dfrac{b}{a} = \cot \alpha$

No matter how large each side of the triangle is in absolute terms, as long as the ratio of the sides relative to each other stays the same, the angle is the same.

4.A.1

(A)

$\sin 34 = \frac{x}{8} \\ x = 8 \sin 34 \approx 4.47$

$\tan 38 = \frac{y}{5} \\ y \approx 3.91$

$\cos 72 = \frac{x}{15} \\ x \approx 4.64$

$\tan 54 = \frac{6}{x} \\ x = \frac{6}{\tan 54} \approx 4.36$

$\cos 48 = \frac{3}{x} \\ x = \frac{3}{\cos 48} \approx 4.48$

$\sin 28 = \frac{4}{y} \\ y \approx 8.52$

(B)

$\tan a = \frac{7}{4} \\ \tan^{-1}(\tan a) = \tan^{-1}(\frac{7}{4}) \\ a = \tan^{-1}(\frac{7}{4}) \\ a \approx 60.3$

$\sin b = \frac{5}{8} \\ b = \sin^{-1}(\frac{5}{8}) \approx 38.7$

$\cos c = \frac{6}{9} \\ c = \cos^{-1}(\frac{6}{9}) \approx 48.2$

$\sin d = \frac{8}{10} \\ d = \sin^{-1}(\frac{8}{10}) \approx 53.1$

Pythagoras' theorem

$A^2 + B^2 = C^2$

set terminal svg size 500,500 enhanced font 'Verdana,10'
unset border

$dataTri << EOD
0 0
3 0
3 2
0 0
EOD

$dataA << EOD
3 0
5 0
5 2
3 2
3 0
EOD

$dataB << EOD
0 0
0 -3
3 -3
3 0
0 0
EOD

$dataC << EOD
0 0
3 2
1 5
-2 3
0 0
EOD

set noxtics
set noytics
set xrange [-3:6]
set yrange [-3.5:5.5]
set nozeroaxis
set xlabel ""
set ylabel ""

set style fill solid 0.1
set label "A" at 2.9,1 right
set label "B" at 1.6,0.1 center
set label "C" at 1.6,0.9 center

plot "$dataTri" with filledcurves notitle, "$dataA" with filledcurves notitle, "$dataB" with filledcurves notitle, "$dataC" with filledcurves notitle

4.A.2

$4^2 + 7^2 = c^2 \\ c = \sqrt{4^2 + 7^2} \approx 8.06$

$5^2 + b^2 = 8^2 \\ b \approx 6.24$

$a^2 + 6^2 = 9^2 \\ a \approx 6.71$

$8^2 + b^2 = 10^2 \\ b = 6$

Do you need to remember all of the trig identities?

Earlier I learned that the difference of two squares is very important to remember, since it is useful in simplifying algebraic expressions and expanding them.

Similarly, there exist many trigonometric identities that are useful.

There are loads of trigonometric identities, but the following are the ones you're most likely to see and use. https://www.purplemath.com/modules/idents.htm

You should memorize a bunch of the common ones and the steps by which they are derived, and you should be comfortable enough with them to derive more obscure ones on your own if asked.

At the bare minimum you should know:

• The pythagorean identity
• formulas for tan, sec, csc, and cot in terms of sin and cos
• formulas for sin(a+b) and cos(a+b)

https://www.reddit.com/r/learnmath/comments/9vqrye/am_i_just_supposed_to_memorize_trig_identities/

Pretty much every trig identity can be derived from $e^{ix} = \cos(x) + i \sin(x)$. However, it is useful to memorize some of the common ones because they will help you a lot in calculus and beyond to quickly identify when an expression can be simplified. https://math.stackexchange.com/a/1341391

My favourite trick: I don't remember any of them. :-) The only thing I have in mind is that this matrix
$\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$
rotates vectors in the plane by an angle θ and matrix multiplication is the same as composition.

https://math.stackexchange.com/a/93362

Basic identities. The following should be memorized as best you can: https://web.archive.org/web/20100709105323/https://math.la.asu.edu/~surgent/mat271/mintrig.pdf [see the last page]

This seems quite a lot to memorize.

Remember: don't try to remember all of these trig identities at once. First see how much you'll actually need them in the future. Computers (SageMath, Wolfram Alpha, etc.) can do a lot of the work for you when solving problems, and if you don't have a computer with you (e.g. when you're doing back-of-the-envelope calculations or just playing around), then maybe you'll get by with only remembering a few identities.

Remember also what Steve Yegge wrote:

The right way to learn math is breadth-first, not depth-first. You need to survey the space, learn the names of things, figure out what's what.

I've found it's much easier to learn and appreciate geometry and trig after you understand what exactly math is — where it came from, where it's going, what it's for. No need to dive right into memorizing geometric proofs and trigonometric identities.

For computer scientists, 95% or more of the interesting math is discrete: i.e., math on the integers.

He also wrote that the following are the most important branches of mathematics for programmers:

• Discrete math
• Probability (which is related to discrete math and set theory)
• Statistics
• Algebra and linear algebra
• Mathematical logic
• Information theory and Kolmogorov complexity

Geometry, trigonometry, differentiation, integration, conic sections, differential equations, and their multidimensional and multivariate versions — these all have important applications. It's just that you don't need to know them right this second. So it probably wasn't a great idea to make you spend years and years doing proofs and exercises with them, was it? If you're going to spend that much time studying math, it ought to be on topics that will remain relevant to you for life.

Properties of triangles

Remember:

• The sum of the angles of a triangle is always 180
• A triangle with two equal sides is called isosceles
• A triangle with three equal sides is called equilateral
• If two triangles have exactly the same angles, they are similar (but not necessarily congruent)
  • All circles are similar to each other
  • All squares are similar to each other
  • All equilateral triangles are similar to each other
• If two triangles have exactly the same sides AND angles, they are congruent. In other words, they can be perfectly overlayed on top of each other (but might need to be rotated or flipped first).

What information do you do need to know of a triangle to be able to tell that two triangles are congruent?

In my mind it's easiest to rephrase the question as "how do you find out the shape of a triangle and the absolute lengths of its sides?". Knowing one of the following should tell me the shape and absolute size of a triangle:

• Two angles and one side
• Two sides and one angle
• Three sides

But checking from Wikipedia, the above is not quite right.

Remember: finding out if two triangles are conguent is more complicated than you might think: https://en.wikipedia.org/wiki/Congruence_(geometry)#Determining_congruence

Even if you know the angle $A$ and the sides $a$ and $b$, you don't know what shape the triangle has, because there are two possibilities for angle $C$ for the same side $a$ length.

To know if two triangles are congruent, you need to know one of these:

• SAS: side-angle-side
• SSS: side-side-side
• ASA: angle-side-angle

In the above triangle I only know side-side-angle (SSA a.k.a. ASS), which isn't by itself enough to tell if two triangles are congruent, but there are some special cases when SSA can be used to tell congruence with some additional information.

The exterior angles of any shape always add up to 360:

The sum of the interior angles depends on the number of sides in the shape. Each new side adds 180 degrees. A triangle is 180 degrees, a square is 360 degrees, etc.

Remember some special cases of right-angled triangles:

• The $90^\circ, 45^\circ, 45^\circ$ triangle (isosceles):
  • Pythagoras' theorem: $1^2 + 1^2 = c^2$
  • $c = \sqrt{2}$
  • $\sin 45 = \frac{a}{c} = \frac{1}{\sqrt{2}}$
  • $\cos 45 = \frac{b}{c} = \frac{1}{\sqrt{2}}$
  • $\tan 45 = \frac{a}{b} = 1$
• The $90^\circ, 60^\circ, 30^\circ$ triangle (i.e. half of an equilateral triangle):
  • Pythagoras' theorem: $a^2 + 1^2 = 2^2$
  • $a = \sqrt{3}$
  • $\sin 60 = \cos 30 = \frac{\sqrt{3}}{2}$
  • $\cos 60 = \sin 30 = \frac{1}{2}$
  • $\tan 60 = \sqrt{3}$
  • $\tan 30 = \frac{1}{\sqrt{3}}$

4.A.h

$\sin a = \dfrac{y}{h} \qquad \cos a = \dfrac{x}{h} \qquad \tan a = \dfrac{y}{x}$

$\sin b = \dfrac{x}{h} \qquad \cos b = \dfrac{y}{h} \qquad \tan b = \dfrac{x}{y}$

Remember these:

$\sin a = \cos b \\ \cos a = \sin b \\ \tan a = \frac{1}{\tan b} \qquad \text{Remember: reciprocal of }p\text{ is }1/p \\ \frac{\sin a}{\cos a} = \tan a$

And especially remember these:

$y^2 + x^2 = h^2 \qquad \text{Pythagoras' theorem} \\ \frac{y^2}{h^2} + \frac{x^2}{h^2} = \frac{h^2}{h^2} = 1 \qquad \text{Divide everything by }h^2 \\ \boxed{\sin^2 a + \cos^2 a = 1}$

And:

$\sin^2 a + \cos^2 a = 1 \\ \frac{\sin^2 a}{\cos^2 a} + \frac{\cos^2 a}{\cos^2 a} = \frac{1}{\cos^2 a} \\ \boxed{\tan^2 a + 1 = \sec^2 a}$

And:

$\sin^2 a + \cos^2 a = 1 \\ \frac{\sin^2 a}{\sin^2 a} + \frac{\cos^2 a}{\sin^2 a} = \frac{1}{\sin^2 a} \\ 1 + \frac{1}{\tan^2 a} = \cosec^2 a \\ \boxed{1 + \cot^2 a = \cosec^2 a}$

Refer to the trigonometric functions at the start of this document.

A thought: $\tan a$ is basically the slope $m$ in the line equation $y = mx + c$.

$y = \tan(a) x + c$

$m = \tan a = \dfrac{y_2 - y_1}{x_2 - x_1}$

$\dfrac{\sin a}{\cos a} = \tan a$

$\sin a = y_2 - y_1 \\ \cos a = x_2 - x_1$

The sine rule, a.k.a. the law of sines

Remember: the sine rule allows you to solve problems of any type of triangle, even if it's not a right-angled triangle, as long as you know at least one angle and its opposite side.

You basically split a triangle into two right-angled triangles, and solve each of them separately with the $\sin$ function.

(It's customary to write the same letter for the angle and its opposite side, e.g. $B$ and $b$.)

$\triangle ABH: \qquad \sin B = \frac{h}{c} \qquad h = c \sin B \\ \triangle ACH: \qquad \sin C = \frac{h}{b} \qquad h = b \sin C$

Therefore:

$h = c \sin B = b \sin C$

Rearrange:

$c \sin B = b \sin C \qquad \dfrac{c \sin B}{\sin C} = b \qquad \dfrac{c}{\sin C} = \dfrac{b}{\sin B}$

The sine rule:

$\boxed{\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}}$

It doesn't seem to be possible to split the triangle into two right-angled triangles in any other way than through $A$. This is more obvious when I draw a flatter triangle:

Drawing a line through $B$ or $C$ doesn't give me two right-angled triangles.

4.B.1

(a)

$C = 180\degree - 78\degree - 65\degree = 37\degree$

$\frac{a}{\sin 78\degree} = \frac{5\text{cm}}{\sin 37\degree} \\ a = \sin 78\degree \frac{5\text{cm}}{\sin 37\degree} \approx 8.13\text{cm}$

$\frac{b}{\sin 65\degree} = \frac{5\text{cm}}{\sin 37\degree} \\ b = \sin 65\degree \frac{5\text{cm}}{\sin 37\degree} \approx 7.53\text{cm}$

(b)

$\frac{6}{\sin A} = \frac{4}{\sin 33\degree} \\ \sin A = \frac{6 \sin 33\degree}{4} \\ A = \sin^{-1}(\frac{6 \sin 33\degree}{4}) \approx 54.8\degree \\ B = 180\degree - A - 33\degree \approx 92.2\degree$

$\frac{b}{\sin B} = \frac{4}{\sin 33\degree} \\ b = \frac{4}{\sin 33\degree} \sin B \\ b = \frac{4}{\sin 33\degree} \sin(180\degree - 33\degree - \sin^{-1}(\frac{6 \sin 33\degree}{4})) \approx 7.34$

Based on those values, my guess about the triangle's shape was wrong. It would look like this instead:

Note: the book says that both triangles are actually correct. A calculator gives the same value for $A = \sin 54.8\degree$ (the "correct" triangle) and $A = \sin 125.2\degree$ (my original triangle). The book will explain later why.

This situation is similar to the earlier situation where you would need more information of the triangle to know its shape. Here I knew side-side-angle (SSA) of the triangle (i.e. $4cm, 6cm, 33\degree$) which isn't enough information to know the shape of the triangle.

(c)

Here I have the same situation as in (b): I know SSA (side-side-angle), which by itself isn't enough to know the shape of the triangle.

But playing around with the triangle, I can't see how it could have any other shape. Wikipedia tells me why:

If two triangles satisfy the SSA condition and the length of the side opposite the angle is greater than or equal to the length of the adjacent side (SSA, or long side-short side-angle), then the two triangles are congruent.

Here the angle is $C$, and the length of the side opposite the angle is $9$, and the length of the adjacent side is $5$. Therefore I can tell the shape of the triangle. (I assume that whenever I can test that two triangles are congruent, I can also tell the exact shape of the two triangles.)

$\frac{5}{\sin A} = \frac{9}{\sin 40\degree} \\ \sin A = \frac{5 \sin 40\degree}{9} \\ A = \sin^{-1}(\frac{5 \sin 40\degree}{9}) \approx 20.9\degree \\ B = 180\degree - 40\degree - A \approx 119.1\degree$

$\frac{b}{\sin 119.078} = \frac{9}{\sin 40\degree} \\ b = \frac{9}{\sin 40\degree} \sin 119.078 \approx 12.24$

The area of a triangle

$\frac{1}{2} ah$

This is intuitive, since the same triangle duplicated always makes a (possibly skewed) rectangle.

But the book also gives another way to calculate the area. This way doesn't require you to calculate $h$ at all!

As seen earlier in the sine rule, $h = c \sin B = b \sin C$.

Therefore a triangle's area can be written as e.g. $\frac{1}{2} a(c \sin B)$, i.e. $\frac{1}{2} ac \sin B$.

In full, the area of a triangle can be written in the following ways:

$\frac{1}{2} ah = \frac{1}{2} ab \sin C = \frac{1}{2} ac \sin B = \frac{1}{2} bc \sin A$

This allows you to simply multiply two sides (e.g. $bc$), halve it, and then multiply that by the sine of the angle that connects those two sides (e.g. $\sin A$).

That works for a right-angled triangle too, since $\sin 90\degree = 1$.

The cosine rule, a.k.a. the law of cosines

If you don't know an angle and its opposite side, then you can't use the sine rule — but you can use the cosine rule.

Start in the same way as with the sine rule: split the triangle into two right-angled triangles, and then solve each of them separately with the $\cos$ function.

$\cos B = \frac{x}{c} \\ x = c \cos B$

$c^2 = h^2 + x^2 \qquad \text{Pythagorean theorem for the left side triangle} \\ x^2 = c^2 - h^2$

$b^2 = h^2 + (a - x)^2 \qquad \text{Pythagorean theorem for the right side triangle} \\ b^2 = h^2 + a^2 - 2ax + x^2 \qquad \text{(Substitute }x^2\text{ with }c^2 - h^2\text{)} \\ b^2 = h^2 + a^2 - 2ax + c^2 - h^2 \qquad \text{(}h^2\text{ cancels out}\text{)} \\ b^2 = a^2 + c^2 - 2ax \qquad \text{(Substitute }x\text{ with }c \cos B\text{)} \\ b^2 = a^2 + c^2 - 2ac \cos B$

The cosine rule:

$\boxed{a^2 = b^2 + c^2 - 2bc \cos A} \\ \boxed{b^2 = a^2 + c^2 - 2ac \cos B} \\ \boxed{c^2 = a^2 + b^2 - 2ab \cos C}$

Remember: the cosine rule should be easy to remember like this:

• When a side is the subject of the equation, the equation always takes the cosine of the side's opposite angle
• The other two sides fill out the rest of the terms in the equation

Things to remember about the sine rule and cosine rule

• Both rules represent the relationship between an angle and its opposite side
• The sine rule can result in ambiguity about the triangle's shape: two different angles can give the same ratio
• The cosine rule can not result in ambiguity, and is therefore safer to use. The cosine of an obtuse angle is negative, and the cosine of an acute angle is positive, which removes any ambiguity.
• The cosine rule is basically the Pythagorean theorem, with a subtraction at the end, e.g. $-2bc \cos A$. With a right-angled triangle, $\cos A = \cos 90\degree = 0$, and the subtraction becomes $0$, and the equation becomes the Pythagorean theorem.

Remember some terminology:

• $90\degree$: right angle
• $\lt 90\degree$: acute angle
• $\gt 90\degree$: obtuse angle

4.B.2

(1a)

$\frac{5}{\sin C} = \frac{7}{\sin B} \qquad \text{Sine rule}$

I don't know $C$ nor $B$, so I'll need to use the cosine rule instead.

$5^2 = 7^2 + 8^2 - 2(7)(8) \cos C \qquad \text{Cosine rule} \\ \cos C = \frac{7^2 + 8^2 - 5^2}{2(7)(8)} \approx 0.786 \\ C = \arccos 0.786 \approx 38.213\degree$

(1b)

$a^2 = 8^2 + 5^2 - 2(8)(5) \cos 72\degree \\ a \approx 8.017$

(1c)

Here I need to solve three unknown angles. I know all three sides. I should be able to use the sine rule here, after I've used the cosine rule to solve one of the angle-and-opposite-side pairs.

Angle B:

$7^2 = 3^2 + 9^2 - 2(3)(9) \cos B \\ \cos B = \frac{3^2 + 9^2 - 7^2}{2(3)(9)} \approx 0.759 \\ B = \arccos 0.759 \approx 40.6\degree$

Now I know the $B$ and $7$ angle-and-opposite-side pair. I should be able to use the sine rule to solve the other pairs.

Angle A:

$\frac{7}{\sin B} = \frac{9}{\sin A} \\ \sin A = \frac{9 \sin 40.601\degree}{7} \approx 0.837 \\ A = \arcsin 0.837 \approx \cancel{56.796} \qquad \text{(}\sin^{-1}\text{ doesn't handle degrees }>90\text{!)}$

$\arcsin$ (i.e. $\sin^{-1}$) only supports degrees $\le 90$. Here we should be getting $123.204$ (i.e. $180\degree - \arcsin 0.837$).

As mentioned earlier, using the cosine rule is safer, since it results in no ambiguity about the shape of the triangle. I'll try:

$9^2 = 3^2 + 7^2 - 2(3)(7) \cos A \\ \cos A = \frac{3^2 + 7^2 - 9^2}{2(3)(7)} \approx -0.548 \\ A = \arccos -0.548 \approx 123.2\degree$

Angle C:

I know the two other angles, so I can simply compute the third one.

$C = 180\degree - 123.2\degree - 40.6\degree = 16.2\degree$

But this I should be able to solve with the sine rule as well, since I know the angle is $\le 90\degree$.

$\frac{7}{\sin B} = \frac{3}{\sin C} \\ \frac{7 \sin C}{\sin B} = 3 \\ \sin C = \frac{3 \sin B}{7} = \frac{3 * \sin 40.601\degree}{7} \approx 0.279 \\ C = \arcsin 0.279 \approx 16.2\degree$

(2a)

An equilateral triangle has all angles at $60\degree$. Therefore $Q = R = \frac{1}{2}60\degree = 30\degree$.

(2b)

A triangle's area: $A = \frac{1}{2}wh$. Therefore $A = \frac{1}{2}(2)(\sqrt{3}) = \sqrt{3}$.

Uhh, scratch that. The book asked "How large is $\angle QPR$", which I misread as asking for the area of $\triangle QPR$. Remember: first understand the problem...

That's a weird way to represent a single angle: $\angle QPR$. Why not just $\angle P$? Anyway, remember to pay attention to this kind of notation.

Answer: the angle $QPR$ (i.e. the angle $P$) is $180\degree - 30\degree - 30\degree = 120\degree$.

(2c)

$(\sqrt{3} + \sqrt{3})^2 = 2^2 + 2^2 - 2(2)(2) \cos P \\ \cos P = \frac{2^2 + 2^2 - (2\sqrt{3})^2}{2(2)(2)} = \frac{4^2 - 2^2(3)}{2(2)(2)} = \frac{2^2 + 2^2 - 2^2(3)}{2(2)(2)} = \frac{2^2(1 + 1 - 1(3))}{2(2)(2)} = -\frac{1}{2} \\ P = \arccos -\frac{1}{2} = 120\degree$

Remember: always try to achieve a closed-form solution (exact value), rather than a numerical solution (approximate value) if possible. And it's probably good to also delay arithmetic, and to factorize and simplify expressions instead, like above. At first I wrote the above solution as an approximate decimal value, but I noticed that my calculation had an error (i.e. the wrong solution). Fixing that error allowed me to find a clean, closed form solution (i.e. an exact fraction value).

(2d)

$\sin Q = \frac{1}{2} \\ \frac{2\sqrt{3}}{\sin P} = \frac{2}{\sin Q} \\ \sin P = \frac{(2\sqrt{3}) \sin Q}{2} = \frac{(2\sqrt{3}) \frac{1}{2}}{2} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \\ P = \arcsin \frac{\sqrt{3}}{2} = 60\degree$

Note how $\sin$ doesn't support degrees $>90$. $P$ should be $120\degree$.

Playing with the trigonometric equations

In my graphs and equations article I came up with the following problem.

set noxtics
set noytics
set xrange [-3:6]
set yrange [-1:5]

$data << EOD
0 0.7
0 3.5
1.4 1.4
EOD

set label "(0, c_a)" at 0,0.7 offset 1.5,0
set label "(0, c_b)" at 0,3.5 offset 1.5,0
set label "(x_a, y_a)" at 1.4,1.4 offset 1.5,0

A(x) = 0.5 * x + 0.7
B(x) = -1.5 * x + 3.5
plot A(x), B(x), "$data" with points lt 6 pt 7 notitle

You know everything about line $A$. About line $B$ you only know that it is perpendicular to line $A$. Lines $A$ and $B$ cross at $(x_a, y_a)$. How do you find out what $c_b$ is for a given value of $x_a$?

Earlier I solved this by using the line equation, and I ended up with $\boxed{c_b = m_a x_a + c_a + \dfrac{x_a}{m_a}}$.

How to solve this with trigonometry?

I see many triangles, two of which are these:

set noxtics
set noytics
set xrange [-3:6]
set yrange [-1:5]

$data << EOD
0 0.7
0 3.5
1.4 1.4
EOD

$data2 << EOD
0 0.7
1.4 0.7
1.4 1.4
EOD

set label "(0, c_a)" at 0,0.7 offset 1,-0.7
set label "(0, c_b)" at 0,3.5 offset 1.5,0
set label "(x_a, y_a)" at 1.4,1.4 offset 1.5,0
set label "h_a" at 0.6,1.3
set label "h_b" at 0.1,2
set label "{/Symbol a}" at 0.4,0.82
set label "{/Symbol b}" at 0.1,3
set style fill solid 0.1

A(x) = 0.5 * x + 0.7
B(x) = -1.5 * x + 3.5
plot A(x), B(x), "$data" with filledcurves lt 2 pt 7 notitle, "$data2" with filledcurves lt 1 pt 7 notitle

$h_a^2 = x_a^2 + (y_a - c_a)^2 \qquad \text{Pythagoras' theorem}$

I realize that $\alpha = \beta$.

To get $h_b$, I need to know at least $\beta$ and $h_a$. The $\sin$ equation uses $\beta$ and $h_a$, so I should be able to rearrange it to get $h_b$.

$\sin \beta = \frac{h_a}{h_b} \\ h_b = \frac{h_a}{\sin \beta} \\ \sin \alpha = \sin \beta \\ \sin \alpha = \frac{y_a - c_a}{h_a} \\ h_b = \frac{h_a}{(y_a - c_a) / h_a} = \frac{h_a^2}{y_a - c_a} = \frac{x_a^2 + (y_a - c_a)^2}{y_a - c_a} = \frac{(y_a - c_a)(x_a^2 / (y_a - c_a) + (y_a - c_a))}{y_a - c_a}$

$\boxed{h_b = \frac{x_a^2}{y_a - c_a} + (y_a - c_a)}$

$c_b = c_a + h_b \\ c_b = c_a + \frac{x_a^2}{y_a - c_a} + (y_a - c_a) \\ \boxed{c_b = \frac{x_a^2}{y_a - c_a} + y_a}$

Testing the equation with the values in the original article:

$c_b = \frac{3^2}{3.5 - 2} + 3.5 = 9.5$

I now see that I could've simplified the original equation further:

$\boxed{c_b = m_a x_a + c_a + \dfrac{x_a}{m_a}}$

Here $m_a = \frac{y_a - c_a}{x_a}$. Therefore we can simplify:

$c_b = y_a - c_a + c_a + \frac{x_a}{m_a} \\ c_b = \frac{x_a}{m_a} + y_a \\ c_b = \frac{x_a}{(y_a - c_a) / x_a} + y_a \\ \boxed{c_b = \frac{x_a^2}{y_a - c_a} + y_a}$

Could I have solved this easier? The two triangles are similar — you can overlay them (with some rotating, flipping and scaling). After you know $h_a$, you could just scale it by the size difference of the two triangles.

set noxtics
set noytics
set xrange [-3:6]
set yrange [-1:5]

$data << EOD
0 0.7
0 3.5
1.4 1.4
EOD

$data2 << EOD
0 0.7
1.4 0.7
1.4 1.4
EOD

set label "(0, c_a)" at 0,0.7 offset 1,-0.7
set label "(0, c_b)" at 0,3.5 offset 1.5,0
set label "(x_a, y_a)" at 1.4,1.4 offset 1.5,0
set label "h_a" at 0.6,1.3
set label "h_b" at 0.1,2
set label "{/Symbol a}" at 0.4,0.82
set label "{/Symbol b}" at 0.1,3
set style fill solid 0.1

A(x) = 0.5 * x + 0.7
B(x) = -1.5 * x + 3.5
plot A(x), B(x), "$data" with filledcurves lt 2 pt 7 notitle, "$data2" with filledcurves lt 1 pt 7 notitle

$\frac{h_a}{h_b} = \frac{y_a - c_a}{h_a} = \sin \alpha = \sin \beta$

$h_b = \frac{h_a^2}{y_a - c_a}$

$c_b = c_a + h_b \\ c_b = c_a + \frac{h_a^2}{y_a - c_a} \\ c_b = c_a + \frac{x^2 + (y_a - c_a)^2}{y_a - c_a} \\ c_b = c_a + \frac{(y_a - c_a)(x^2 / (y_a - c_a) + (y_a - c_a))}{y_a - c_a} \\ c_b = c_a + x^2 / (y_a - c_a) + (y_a - c_a) \\ \boxed{c_b = \frac{x^2}{y_a - c_a} + y_a}$

This approach (inspired by understanding that the two triangles are similar) was pretty much the same as the earlier one, but this one felt a little bit more controlled or intuitive. Each term was successively decomposed into a more complicated expression, and the whole thing was then simplified.

The unit circle

Trigonometry

You think it's about triangles, but really it's about circles.

set terminal svg size 408,400 enhanced font 'Verdana,10'

set xrange [-1.2:1.2]
set yrange [-1.2:1.2]
set xtics 1
set ytics 1

set xlabel ""
set ylabel ""
set noborder
set object circle at 0,0 radius 1 lw 0.5 fc rgb "black"
set style fill solid 0.1

$triangle << EOD
0 0
0.707 0.707
0.707 0
EOD

plot $triangle with filledcurves notitle

See https://www.geogebra.org/m/keqhdkaj.

Remember the equation of the circle (that is centered on the origin): $x^2 + y^2 = r^2$

If the circle is not centered on the origin, but instead on point $(a,b)$, you need to offset it. The equation becomes $(x - a)^2 + (y - b)^2 = r^2$, or:

$(x^2 - 2ax + a^2) + (y^2 - 2by + b^2) = r^2$

In the unit circle $r^2 = 1$. Let's see what value of $y$ I get for specific values of $x$:

$x^2 + y^2 = 1 \\ y^2 = 1 - x^2 \\ f(x) = \sqrt{1 - x^2}$

$f(1) = 0 \\ f(0) = 1 \\ f(-1) = 0 \\ f(0.5) = 0.866 \qquad \text{!!} \\ f(0.707) \approx 0.707 \qquad \text{??}$

set terminal svg size 408,400 enhanced font 'Verdana,10'

set xrange [-1.2:1.2]
set yrange [-1.2:1.2]
set xtics 1
set ytics 1

set xlabel ""
set ylabel ""
set noborder
set object circle at 0,0 radius 1 lw 0.5 fc rgb "black"
set style fill solid 0.1

$data << EOD
1 0
0 1
-1 0
0.5 0.866
0.707 0.707
EOD

plot $data with points pt 7 notitle

Why $0.707$? The circle's equation is $x^2 + y^2 = r^2$. In a unit circle that's $x^2 + y^2 = 1$. If you want $x$ and $y$ to be equal (to get exactly $\frac{1}{8}$ of a turn), then $x^2 = y^2$, so we have $x^2 + x^2 = 2x^2 = 1$. Solve for $x$:

$2x^2 = 1 \\ x^2 = \frac{1}{2} \\ x = \sqrt{\frac{1}{2}} \approx 0.707$

At this point I became a bit demotivated with trigonometry and/or the current book. I decided to switch to Stitz & Zeager's precalculus book.