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
θ°adjopphyp

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.

Opposite, adjacent and hypotenuse are named from the angle.

Concept
Definitions from a right triangle

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.

Example
Example: all six ratios of a 3-4-5 triangle

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.

Example
Example: find the missing side first

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}.

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).

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.

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

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

  • 2pt
  • 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

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.

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.

Cofunctions and the Complementary-Angle Relationship

Concept
Cofunction identities

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.

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.

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).

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

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∘≈50(0.8391)≈41.96 ft.h=50\tan 40^\circ\approx 50(0.8391)\approx 41.96\text{ ft.}
Example
Example: angle of depression

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.

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.

  • 2pt
  • 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)

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

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

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

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:

  • 2pt
  • 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.

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.