Right Triangle Trigonometry

Study Sheet

Right Triangle Trigonometry

The six ratios, special angles, solving triangles, cofunctions, and applications

The Six Trigonometric Ratios (SOH-CAH-TOA)

Tip

Rounding conventions used in this topic: unless a problem asks for an exact value, round side lengths to the nearest hundredth (0.010.01) and angle measures to the nearest hundredth of a degree (0.01∘0.01^\circ). Keep full precision in your calculator until the final step, and make sure your calculator is in degree mode.

Sine, cosine and tangent are ratios of sides in a right triangle: opposite over hypotenuse, adjacent over hypotenuse, opposite over adjacent (SOH-CAH-TOA). They let one angle and one side determine the rest of the triangle, and their inverses recover an angle from a ratio. The reciprocal functions and cofunction identities round out the set.

Concept
Definitions from a right triangle
oppositeadjacenthypotenuseθ
Opposite and adjacent are named from θ\theta; the hypotenuse faces the right angle.

For an acute angle θ\theta in a right triangle, label the sides relative to θ\theta: the opposite leg (across from θ\theta), the adjacent leg (next to θ\theta, not the hypotenuse), and the hypotenuse (across from the right angle). Then

sin⁡θ=opphyp,cos⁡θ=adjhyp,tan⁡θ=oppadj\sin\theta=\frac{\text{opp}}{\text{hyp}},\qquad \cos\theta=\frac{\text{adj}}{\text{hyp}},\qquad \tan\theta=\frac{\text{opp}}{\text{adj}}
csc⁡θ=hypopp,sec⁡θ=hypadj,cot⁡θ=adjopp\csc\theta=\frac{\text{hyp}}{\text{opp}},\qquad \sec\theta=\frac{\text{hyp}}{\text{adj}},\qquad \cot\theta=\frac{\text{adj}}{\text{opp}}

Remember the mnemonic SOH-CAH-TOA: Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj.

Why should a ratio of sides depend only on the angle? Any two right triangles with the same acute angle θ\theta have all three angles equal, so they are similar: one is a scaled copy of the other, and scaling multiplies every side by the same factor, leaving every ratio unchanged. So opphyp\dfrac{\text{opp}}{\text{hyp}} is a property of θ\theta itself, not of the particular triangle, and it deserves a name.

Example
Example: all six ratios of a 3-4-5 triangle
345θ
Side 33 is opposite θ\theta, side 44 is adjacent to it.

A right triangle has legs 33 (opposite θ\theta) and 44 (adjacent to θ\theta) and hypotenuse 55.

sin⁡θ=35,cos⁡θ=45,tan⁡θ=34\sin\theta=\tfrac{3}{5},\quad \cos\theta=\tfrac{4}{5},\quad \tan\theta=\tfrac{3}{4}
csc⁡θ=53,sec⁡θ=54,cot⁡θ=43\csc\theta=\tfrac{5}{3},\quad \sec\theta=\tfrac{5}{4},\quad \cot\theta=\tfrac{4}{3}

The three ratios on the bottom row are the reciprocals of the top row.

Side note
Where it goes wrong

Mixing up "opposite" and "adjacent" — they depend on WHICH acute angle you are using. Label the sides relative to the angle in the problem before writing a ratio.

Example
Example: find the missing side first
51213θ
The adjacent leg 1212 comes from 132−52\sqrt{13^2-5^2}.

A right triangle has opposite =5=5 and hypotenuse =13=13. Find cos⁡θ\cos\theta. By the Pythagorean theorem the adjacent side is 132−52=144=12\sqrt{13^2-5^2}=\sqrt{144}=12, so cos⁡θ=1213\cos\theta=\dfrac{12}{13} and tan⁡θ=512\tan\theta=\dfrac{5}{12}.

Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2
Tip

Reciprocal identities: csc⁡θ=1sin⁡θ\csc\theta=\dfrac{1}{\sin\theta},  sec⁡θ=1cos⁡θ\sec\theta=\dfrac{1}{\cos\theta},  cot⁡θ=1tan⁡θ\cot\theta=\dfrac{1}{\tan\theta}. Also tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta}.

Try it
Quick check

A right triangle has legs 88 and 1515, and θ\theta is the acute angle opposite the leg 88. Find all six trigonometric ratios of θ\theta.

Show the answer

Answer: the hypotenuse is 82+152=289=17\sqrt{8^2+15^2}=\sqrt{289}=17, so sin⁡θ=817\sin\theta=\dfrac{8}{17}, cos⁡θ=1517\cos\theta=\dfrac{15}{17}, tan⁡θ=815\tan\theta=\dfrac{8}{15}, csc⁡θ=178\csc\theta=\dfrac{17}{8}, sec⁡θ=1715\sec\theta=\dfrac{17}{15}, cot⁡θ=158\cot\theta=\dfrac{15}{8}. The side 88 is opposite θ\theta and 1515 is adjacent to it; the bottom three are the top three flipped.

Exact Values of the Special Angles 30∘,45∘,60∘30^\circ,45^\circ,60^\circ

Concept
The two special triangles

Every special-angle value comes from two triangles. The 45∘45^\circ-45∘45^\circ-90∘90^\circ triangle has legs 1,11,1 and hypotenuse 2\sqrt{2}. The 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle has sides 11 (short leg, opposite 30∘30^\circ), 3\sqrt{3} (long leg, opposite 60∘60^\circ), and 22 (hypotenuse).

Both triangles are halves of familiar shapes. Cut a square of side 11 along its diagonal: each half has legs 1,11,1, two 45∘45^\circ angles, and hypotenuse 12+12=2\sqrt{1^2+1^2}=\sqrt2. Cut an equilateral triangle of side 22 down its altitude: each half has hypotenuse 22, short leg 11 (half the base), angles 30∘30^\circ and 60∘60^\circ, and long leg 22−12=3\sqrt{2^2-1^2}=\sqrt3. Every entry in the table is a ratio of two of these sides.

Half a square (left) and half an equilateral triangle (right).

Concept
Table of exact values
θsin⁡θcos⁡θtan⁡θ30∘12323345∘2222160∘32123\begin{array}{c|ccc} \theta & \sin\theta & \cos\theta & \tan\theta\\ \hline 30^\circ & \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} & \dfrac{\sqrt{3}}{3}\\[6pt] 45^\circ & \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2} & 1\\[6pt] 60^\circ & \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} & \sqrt{3} \end{array}

Reciprocal functions come from flipping these: e.g. csc⁡30∘=2\csc 30^\circ=2, sec⁡60∘=2\sec 60^\circ=2, cot⁡45∘=1\cot 45^\circ=1.

Example
Example: exact evaluation

Evaluate sin⁡30∘cos⁡60∘+cos⁡30∘sin⁡60∘\sin 30^\circ\cos 60^\circ+\cos 30^\circ\sin 60^\circ.

(12) ⁣(12)+(32) ⁣(32)=14+34=1.\left(\tfrac{1}{2}\right)\!\left(\tfrac{1}{2}\right)+\left(\tfrac{\sqrt{3}}{2}\right)\!\left(\tfrac{\sqrt{3}}{2}\right)=\tfrac{1}{4}+\tfrac{3}{4}=1.
Tip

Tip: tan⁡30∘=13\tan 30^\circ=\dfrac{1}{\sqrt3} is usually written with a rational denominator as 33\dfrac{\sqrt3}{3}. As θ\theta grows from 0∘0^\circ to 90∘90^\circ, sin⁡θ\sin\theta increases and cos⁡θ\cos\theta decreases, so their values “cross” at 45∘45^\circ.

Try it
Quick check

Evaluate exactly: 2sin⁡45∘cos⁡45∘+tan⁡260∘2\sin 45^\circ\cos 45^\circ+\tan^2 60^\circ.

Show the answer

Answer: 44. From the 45∘45^\circ triangle, sin⁡45∘=cos⁡45∘=22\sin 45^\circ=\cos 45^\circ=\dfrac{\sqrt2}{2}, so 2⋅22⋅22=2⋅24=12\cdot\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt2}{2}=2\cdot\dfrac{2}{4}=1. From the 3030-6060-9090 triangle, tan⁡60∘=31\tan 60^\circ=\dfrac{\sqrt3}{1}, so tan⁡260∘=3\tan^2 60^\circ=3. Total: 1+3=41+3=4.

Calculator Values and Solving Right Triangles

Concept
Using a calculator

For values that are not special angles, use your calculator (in degree mode). To find a side, multiply by a trig value; to find an angle, use the inverse keys sin⁡−1,cos⁡−1,tan⁡−1\sin^{-1},\cos^{-1},\tan^{-1}. Solving a right triangle means finding all three sides and all three angles. You need one side plus either one more side or one acute angle.

Example
Example: given an angle and the hypotenuse
ABCa ≈ 5.74b ≈ 8.19c = 1035°55°
Opposite 35∘35^\circ is csin⁡35∘c\sin 35^\circ; adjacent is ccos⁡35∘c\cos 35^\circ.

In a right triangle the right angle is at CC; angle A=35∘A=35^\circ and hypotenuse c=10c=10.

  • B=90∘−35∘=55∘B=90^\circ-35^\circ=55^\circ.
  • a=csin⁡A=10sin⁡35∘≈5.74a=c\sin A=10\sin 35^\circ\approx 5.74 (opposite AA).
  • b=ccos⁡A=10cos⁡35∘≈8.19b=c\cos A=10\cos 35^\circ\approx 8.19 (adjacent to AA).
Example
Example: given two sides, find an angle
ABCa = 9b = 12c = 1536.87°53.13°
Both legs are known, so tan⁡A=912\tan A=\tfrac{9}{12}.

A right triangle has legs a=9a=9 and b=12b=12. The hypotenuse is c=92+122=225=15c=\sqrt{9^2+12^2}=\sqrt{225}=15. Then tan⁡A=912=0.75\tan A=\dfrac{9}{12}=0.75, so A=tan⁡−1(0.75)≈36.87∘A=\tan^{-1}(0.75)\approx 36.87^\circ and B≈53.13∘B\approx 53.13^\circ.

Side note
Where it goes wrong

A calculator in radian mode. Typing 10sin⁡3510\sin 35 in radian mode gives 10sin⁡(35 rad)≈−4.2810\sin(35\text{ rad})\approx -4.28, a negative side length, because 3535 radians is more than five full turns. A negative or wildly off length is the signal: switch to degree mode, where 10sin⁡35∘≈5.7410\sin 35^\circ\approx 5.74.

Tip

Tip: choose the ratio that uses the side you are given and the side you want. Use sin⁡/csc⁡\sin/\csc with opposite & hypotenuse, cos⁡/sec⁡\cos/\sec with adjacent & hypotenuse, tan⁡/cot⁡\tan/\cot with the two legs.

Try it
Quick check

Right triangle ABCABC has its right angle at CC, B=62∘B=62^\circ, and leg a=BC=7a=BC=7. Find the other leg bb and the hypotenuse cc.

Show the answer

Answer: b≈13.17b\approx 13.17, c≈14.91c\approx 14.91. Relative to BB, the leg a=7a=7 is adjacent and bb is opposite. So tan⁡62∘=b7\tan 62^\circ=\dfrac{b}{7} gives b=7tan⁡62∘≈13.17b=7\tan 62^\circ\approx 13.17, and cos⁡62∘=7c\cos 62^\circ=\dfrac{7}{c} gives c=7cos⁡62∘≈14.91c=\dfrac{7}{\cos 62^\circ}\approx 14.91. Check with Pythagoras: 72+13.172≈14.91\sqrt{7^2+13.17^2}\approx 14.91.

Cofunctions and the Complementary-Angle Relationship

Concept
Cofunction identities
abcθ90° − θ
Leg aa is opposite θ\theta and adjacent to 90∘−θ90^\circ-\theta, so sin⁡θ=ac=cos⁡(90∘−θ)\sin\theta=\tfrac ac=\cos(90^\circ-\theta).

In a right triangle the two acute angles are complementary (they sum to 90∘90^\circ). The side opposite one angle is adjacent to the other, so

sin⁡θ=cos⁡(90∘−θ),tan⁡θ=cot⁡(90∘−θ),sec⁡θ=csc⁡(90∘−θ),\sin\theta=\cos(90^\circ-\theta),\qquad \tan\theta=\cot(90^\circ-\theta),\qquad \sec\theta=\csc(90^\circ-\theta),

and the same with the pairs reversed. “Co” functions (cosine, cotangent, cosecant) are the functions of the complementary angle.

The two acute angles share one triangle, and the leg opposite θ\theta is the leg adjacent to 90∘−θ90^\circ-\theta. So the fraction that leghyp\dfrac{\text{that leg}}{\text{hyp}} is sin⁡θ\sin\theta when read from one corner and cos⁡(90∘−θ)\cos(90^\circ-\theta) when read from the other: the same number with two names. The prefix co- is short for complement: cosine is the sine of the complementary angle.

Example
Example: rewrite and solve

Since sin⁡20∘=cos⁡70∘\sin 20^\circ=\cos 70^\circ, we can fill in sin⁡20∘=cos⁡( 70∘ ‾)\sin 20^\circ=\cos(\underline{\ 70^\circ\ }). To solve sin⁡(x)=cos⁡(x+10∘)\sin(x)=\cos(x+10^\circ), set the angles complementary: x+(x+10∘)=90∘x+(x+10^\circ)=90^\circ, so 2x=80∘2x=80^\circ and x=40∘x=40^\circ.

Tip

Tip: a cofunction equation such as sin⁡A=cos⁡B\sin A=\cos B is true exactly when A+B=90∘A+B=90^\circ. This lets you solve for an unknown angle without a calculator.

Try it
Quick check

Solve for the acute angle xx: tan⁡(2x+5∘)=cot⁡(x+10∘)\tan(2x+5^\circ)=\cot(x+10^\circ).

Show the answer

Answer: x=25∘x=25^\circ. tan⁡A=cot⁡B\tan A=\cot B exactly when AA and BB are complementary (for acute angles), so (2x+5∘)+(x+10∘)=90∘(2x+5^\circ)+(x+10^\circ)=90^\circ, giving 3x=75∘3x=75^\circ and x=25∘x=25^\circ. Check: tan⁡55∘=cot⁡35∘\tan 55^\circ=\cot 35^\circ, and 55+35=9055+35=90.

Applications: Elevation, Depression, and Bearings

Concept
Angles of elevation and depression

An angle of elevation is measured upward from a horizontal line to a line of sight; an angle of depression is measured downward from a horizontal line to a line of sight. Because the two horizontal lines are parallel, the angle of depression from the top equals the angle of elevation from the bottom (alternate interior angles).

Both angles are measured from a horizontal, never from the vertical wall or cliff. The horizontal through the observer at the top and the ground line at the bottom are parallel, and the line of sight is a transversal, so the angle of depression and the angle of elevation are alternate interior angles and are equal. That is why a depression problem can be solved inside the ground-level right triangle.

Concept
Bearings

A bearing such as N60∘E\text{N}60^\circ\text{E} names a direction by an acute angle measured from the north-south line toward the east or west. Resolve a displacement of length dd on bearing NαE\text{N}\alpha\text{E} into a north component dcos⁡αd\cos\alpha and an east component dsin⁡αd\sin\alpha.

Example
Example: angle of elevation
50 fth ≈ 41.95 ft40°
The height is opposite the 40∘40^\circ angle; the 5050 ft is adjacent.

From a point 5050 ft from the base of a building, the angle of elevation to the top is 40∘40^\circ. The height is

h=50tan⁡40∘≈41.95 ft.h=50\tan 40^\circ\approx 41.95\text{ ft.}
Example
Example: angle of depression
35°35°30 mx ≈ 42.84 mhorizontal
Depression at the top equals elevation at the boat (alternate interior angles).

From the top of a 3030 m cliff, the angle of depression to a boat is 35∘35^\circ. The horizontal distance from the base of the cliff to the boat is

x=30tan⁡35∘≈300.7002≈42.84 m.x=\frac{30}{\tan 35^\circ}\approx\frac{30}{0.7002}\approx 42.84\text{ m.}
Tip

Tip for two-triangle problems: draw one picture, label the unknown height hh and set up one equation per triangle. Often both equations contain hh; subtract or substitute to eliminate it.

Try it
Quick check

A kite string 120120 m long is pulled straight and makes a 52∘52^\circ angle with the ground. How high is the kite?

Show the answer

Answer: about 94.5694.56 m. The string is the hypotenuse and the height is opposite the 52∘52^\circ angle, so sin⁡52∘=h120\sin 52^\circ=\dfrac{h}{120} and h=120sin⁡52∘≈94.56h=120\sin 52^\circ\approx 94.56 m.

Going Deeper: Advanced Right-Triangle Trig

Tip

How to work these: advanced problems almost always hide two right triangles that share a side. Draw one clean picture, name every point, and write one equation per triangle. Keep exact values (simplified radicals) until the last step, then round lengths to 0.010.01 and angles to 0.01∘0.01^\circ.

Concept
Two observers / two towers (shared height)

When two observers on the same horizontal line sight the same point, both right triangles share the vertical height hh. Write each horizontal distance in terms of hh using tan⁡\tan, then use the known separation to eliminate hh.

  • Observers on opposite sides, separation DD, elevations α\alpha and β\beta: @@BLOCK0@@
  • Observers on the same side, one behind the other by DD: @@BLOCK1@@

Factor out hh and divide. The subtraction case is the usual “walk toward the tower” setup.

Example
Example: a tower between two observers (exact then rounded)
30°45°ABhh/tan 30°h/tan 45°300 m
The two ground pieces are each written in terms of hh and add to 300300.

A vertical tower stands between two observers AA and BB who are 300300 m apart on level ground. From AA the angle of elevation to the top is 30∘30^\circ; from BB it is 45∘45^\circ. Find the height hh.

htan⁡30∘+htan⁡45∘=300  ⟹  h3+h=300  ⟹  h(3+1)=300.\frac{h}{\tan 30^\circ}+\frac{h}{\tan 45^\circ}=300 \;\Longrightarrow\; h\sqrt{3}+h=300 \;\Longrightarrow\; h(\sqrt3+1)=300.

Rationalize by multiplying by 3−13−1\dfrac{\sqrt3-1}{\sqrt3-1}:

h=3003+1=300(3−1)(3+1)(3−1)=300(3−1)2=150(3−1)≈109.81 m.h=\frac{300}{\sqrt3+1}=\frac{300(\sqrt3-1)}{(\sqrt3+1)(\sqrt3-1)}=\frac{300(\sqrt3-1)}{2}=150(\sqrt3-1)\approx 109.81\text{ m}.
Concept
3D right-triangle problems

A three-dimensional figure is solved by finding a sequence of right triangles that lie in different planes but share an edge. A common tool is the space diagonal of a box with edges a,b,ca,b,c: first the base diagonal is d=a2+b2d=\sqrt{a^2+b^2}, then the space diagonal is d2+c2=a2+b2+c2\sqrt{d^2+c^2}=\sqrt{a^2+b^2+c^2}. The angle the space diagonal makes with the base satisfies tan⁡ϕ=cd=ca2+b2\tan\phi=\dfrac{c}{d}=\dfrac{c}{\sqrt{a^2+b^2}}.

Example
Example: angle of a box's space diagonal with its base
φ685
The base diagonal (1010) and the height (55) are the legs of the right triangle that holds φ\varphi.

A rectangular box has base edges a=6a=6 and b=8b=8 and height c=5c=5. The base diagonal is

d=62+82=100=10.d=\sqrt{6^2+8^2}=\sqrt{100}=10.

The space diagonal is 102+52=125=55≈11.18\sqrt{10^2+5^2}=\sqrt{125}=5\sqrt5\approx 11.18. The angle ϕ\phi it makes with the base is

ϕ=tan⁡−1 ⁣cd=tan⁡−1510=tan⁡−1(0.5)≈26.57∘.\phi=\tan^{-1}\!\frac{c}{d}=\tan^{-1}\frac{5}{10}=\tan^{-1}(0.5)\approx 26.57^\circ.
Concept
The altitude-on-hypotenuse geometric mean

Drop the altitude of length hh from the right angle to the hypotenuse. It splits the hypotenuse into segments pp and qq and creates two smaller triangles, each similar to the original. This gives three geometric-mean relations:

h=pq,(leg1)=p (p+q),(leg2)=q (p+q).h=\sqrt{pq},\qquad (\text{leg}_1)=\sqrt{p\,(p+q)},\qquad (\text{leg}_2)=\sqrt{q\,(p+q)}.

In words: the altitude is the geometric mean of the two hypotenuse pieces, and each leg is the geometric mean of the whole hypotenuse and the piece adjacent to that leg.

Example
Example: geometric mean on the hypotenuse
p = 4q = 9h = 62√133√13
The altitude splits the hypotenuse 1313 into 44 and 99; h2=4⋅9h^2=4\cdot 9.

The altitude from the right angle meets the hypotenuse, cutting it into pieces p=4p=4 and q=9q=9. Then

h=4⋅9=36=6,leg1=4⋅13=213≈7.21,leg2=9⋅13=313≈10.82.h=\sqrt{4\cdot 9}=\sqrt{36}=6,\qquad \text{leg}_1=\sqrt{4\cdot 13}=2\sqrt{13}\approx 7.21,\qquad \text{leg}_2=\sqrt{9\cdot 13}=3\sqrt{13}\approx 10.82.

Check: leg12+leg22=52+117=169=132\text{leg}_1^2+\text{leg}_2^2=52+117=169=13^2, the square of the full hypotenuse p+q=13p+q=13.

Concept
Bearings and navigation with two legs

Give each leg of a trip as a bearing (measured from north) and a length. Resolve every leg into north and east components (dcos⁡d\cos for the N-S part, dsin⁡d\sin for the E-W part). When two consecutive bearings differ by exactly 90∘90^\circ, the two legs are perpendicular, so the start-to-finish distance is just d12+d22\sqrt{d_1^2+d_2^2} and the turn angle is tan⁡−1(d2/d1)\tan^{-1}(d_2/d_1) measured from the first leg.

Example
Example: two perpendicular legs
N40°12 km8 km≈ 14.42 kmO
Bearings 040∘040^\circ and 130∘130^\circ differ by 90∘90^\circ, so the legs meet at a right angle.

A ship sails 1212 km on bearing N40∘E\text{N}40^\circ\text{E}, then turns and sails 88 km on bearing S50∘E\text{S}50^\circ\text{E}. The two bearings are 040∘040^\circ and 130∘130^\circ, which differ by 90∘90^\circ, so the legs are perpendicular. The direct distance from start to finish is

122+82=208=413≈14.42 km.\sqrt{12^2+8^2}=\sqrt{208}=4\sqrt{13}\approx 14.42\text{ km}.

The course swings clockwise from the first leg by tan⁡−1 ⁣812=tan⁡−1(0.66‾)≈33.69∘\tan^{-1}\!\dfrac{8}{12}=\tan^{-1}(0.6\overline{6})\approx 33.69^\circ, so the final bearing from the start is about 40∘+33.69∘=73.69∘40^\circ+33.69^\circ=73.69^\circ, i.e. N73.69∘E\text{N}73.69^\circ\text{E}.

Concept
Right triangles inside inscribed figures

Two facts turn geometry problems into right-triangle trig:

  • Thales' theorem: any triangle inscribed in a semicircle, with the diameter as one side, has its right angle on the circle. So a diameter-based inscribed triangle is automatically right.
  • Regular nn-gon in a circle of radius RR: draw radii to two adjacent vertices and the apothem to the midpoint of that side. The half-central-angle is 180∘n\dfrac{180^\circ}{n}, giving @@BLOCK0@@
Concept
Optimizing a viewing angle

A picture (or screen) hangs on a wall with its bottom edge aa above eye level and its top edge bb above eye level. A viewer standing xx from the wall sees the picture within the viewing angle

θ(x)=tan⁡−1bx−tan⁡−1ax.\theta(x)=\tan^{-1}\frac{b}{x}-\tan^{-1}\frac{a}{x}.

The angle is small when xx is very small (you look almost straight up) or very large (the picture shrinks), so a best distance sits in between. Calculus shows θ\theta is largest at

x=ab,x=\sqrt{ab},

the geometric mean of the two heights. This is another place the geometric mean appears in right-triangle work.

Tip

Exact-value chains and cofunction tricks: cofunctions collapse long products. Because tan⁡θ tan⁡(90∘−θ)=tan⁡θ cot⁡θ=1\tan\theta\,\tan(90^\circ-\theta)=\tan\theta\,\cot\theta=1, a symmetric product telescopes, e.g.

tan⁡1∘ tan⁡2∘⋯tan⁡89∘=1,\tan 1^\circ\,\tan 2^\circ\cdots\tan 89^\circ=1,

by pairing each factor tan⁡k∘\tan k^\circ with tan⁡(90∘−k∘)\tan(90^\circ-k^\circ); the lone middle term tan⁡45∘=1\tan 45^\circ=1. Likewise sin⁡2θ+sin⁡2(90∘−θ)=sin⁡2θ+cos⁡2θ=1\sin^2\theta+\sin^2(90^\circ-\theta)=\sin^2\theta+\cos^2\theta=1. Look for complementary pairs before reaching for a calculator.

Try it
Quick check

From a point on level ground the angle of elevation to the top of a tower is 30∘30^\circ. After walking 4040 m straight toward the tower, it is 60∘60^\circ. Find the tower's height exactly.

Show the answer

Answer: h=203≈34.64h=20\sqrt3\approx 34.64 m. Both right triangles share the height hh. The two ground distances are htan⁡30∘=h3\dfrac{h}{\tan 30^\circ}=h\sqrt3 and htan⁡60∘=h3\dfrac{h}{\tan 60^\circ}=\dfrac{h}{\sqrt3}, and they differ by the 4040 m walked: h3−h3=40h\sqrt3-\dfrac{h}{\sqrt3}=40. The left side is h⋅3−13=2h3h\cdot\dfrac{3-1}{\sqrt3}=\dfrac{2h}{\sqrt3}, so h=203h=20\sqrt3.

Why It Works

The ratios are functions of the angle alone because any two right triangles sharing an acute angle are similar (AA), so their corresponding side ratios agree. Cofunction identities like sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ - \theta) = \cos\theta hold because the two acute angles of a right triangle are complementary and swap the roles of opposite and adjacent. The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is a2+b2=c2a^2 + b^2 = c^2 divided by c2c^2.

Reminder — The Pythagorean identity:  sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Special right triangles
45∘-45∘-90∘: 1:1:230∘-60∘-90∘: 1:3:245^\circ\text{-}45^\circ\text{-}90^\circ:\ 1:1:\sqrt2 \qquad 30^\circ\text{-}60^\circ\text{-}90^\circ:\ 1:\sqrt3:2

What it means. Two triangles whose sides you can write down without a calculator.

Example. A 4545-4545-9090 triangle with legs 55 has hypotenuse 525\sqrt2.

Why it works. A 4545-4545-9090 is half a square cut along its diagonal, so the legs match and Pythagoras gives hypotenuse 2\sqrt2. A 3030-6060-9090 is half an equilateral triangle: the hypotenuse is a full side 22, the short leg is half a side 11, and the long leg is 22−12=3\sqrt{2^2-1^2}=\sqrt3.

Tip. The short leg is always opposite the 30∘30^\circ angle. Match sides to angles before assigning 11, 3\sqrt3, 22.

Tip
The trigonometric ratios (SOH-CAH-TOA)
sin⁡θ=opphyp,cos⁡θ=adjhyp,tan⁡θ=oppadj\sin\theta=\frac{\text{opp}}{\text{hyp}},\qquad \cos\theta=\frac{\text{adj}}{\text{hyp}},\qquad \tan\theta=\frac{\text{opp}}{\text{adj}}

What it means. Ratios of sides in a right triangle that depend only on the angle.

Example. In a 33-44-55 right triangle, sin⁡=35\sin=\tfrac35, cos⁡=45\cos=\tfrac45, tan⁡=34\tan=\tfrac34.

Why it works. Any two right triangles with the same acute angle are similar, so matching side ratios are equal. That makes each ratio a function of the angle alone — which is what lets a table or calculator store them.

Tip. tan⁡θ=sin⁡θcos⁡θ\tan\theta=\tfrac{\sin\theta}{\cos\theta}, since dividing opp/hyp by adj/hyp cancels the hypotenuse.