Right Triangles & Trig

Study Sheet

Right Triangles & Trig

Pythagoras, special triangles, SOH-CAH-TOA, and angles of elevation

The Pythagorean Theorem

Concept
The Pythagorean Theorem
abc

In any right triangle, the two shorter sides are the legs (aa and bb) and the longest side --- always across from the right angle --- is the hypotenuse (cc). The theorem says:

a2+b2=c2.a^2 + b^2 = c^2.

The two legs squared and added always equal the hypotenuse squared. Because cc is the biggest side, it sits alone on its own side of the equation.

The legs and the hypotenuse.

Reminder — The Pythagorean theorem:a2+b2=c2a^2+b^2=c^2
Example
Finding the hypotenuse

The legs are 66 and 88. Find the hypotenuse cc.

c2=62+82=36+64=100,c=100=10.c^2 = 6^2 + 8^2 = 36 + 64 = 100, \qquad c = \sqrt{100} = 10.
Example
Finding a leg

The hypotenuse is 1313 and one leg is 55. Find the other leg bb. The hypotenuse stays alone:

52+b2=132    25+b2=169    b2=144    b=12.5^2 + b^2 = 13^2 \;\Rightarrow\; 25 + b^2 = 169 \;\Rightarrow\; b^2 = 144 \;\Rightarrow\; b = 12.
Tip

Tip: To find a leg, subtract: b=c2a2b = \sqrt{c^2 - a^2}. To find the hypotenuse, add: c=a2+b2c = \sqrt{a^2+b^2}. If you ever get the hypotenuse smaller than a leg, you added when you should have subtracted.

The Converse: Right, Acute, or Obtuse?

Concept
Converse of the Pythagorean Theorem
55°

Take the three sides of a triangle and let cc be the longest. Compare a2+b2a^2+b^2 with c2c^2:

  • [leftmargin=6mm]
  • If a2+b2=c2a^2 + b^2 = c^2, the triangle is right.
  • If a2+b2>c2a^2 + b^2 > c^2, the triangle is acute (all angles under 9090^\circ).
  • If a2+b2<c2a^2 + b^2 < c^2, the triangle is obtuse (one angle over 9090^\circ).

A bigger c2c^2 “pulls” the longest side apart into an obtuse angle.

An angle measures a turn.

Example
Classifying a triangle

Sides 77, 88, 1212. The longest is 1212, so

72+82=49+64=113,122=144.7^2 + 8^2 = 49 + 64 = 113, \qquad 12^2 = 144.

Since 113<144113 < 144, we have a2+b2<c2a^2+b^2 < c^2, so the triangle is obtuse.

Tip

Remember: Always square the longest side for c2c^2. “Sum of squares of the two short sides” versus “square of the long side” tells the whole story.

Pythagorean Triples

Concept
Whole-number right triangles
55°

A Pythagorean triple is a set of three whole numbers that fit a2+b2=c2a^2+b^2=c^2 exactly. Memorizing a few lets you skip the arithmetic. The common ones are:

3-4-5,5-12-13,8-15-17,7-24-25.3\text{-}4\text{-}5, \qquad 5\text{-}12\text{-}13, \qquad 8\text{-}15\text{-}17, \qquad 7\text{-}24\text{-}25.

Any multiple of a triple is also a triple. Multiply 33-44-55 by 22 to get 66-88-1010; by 33 to get 99-1212-1515; by 1010 to get 3030-4040-5050.

An angle measures a turn.

Example
Spotting a triple

The legs are 1010 and 2424. These are 2×52\times 5 and 2×122\times 12, so this is the 55-1212-1313 triple doubled. The hypotenuse is 2×13=262 \times 13 = 26. Check: 102+242=100+576=676=26210^2+24^2 = 100+576 = 676 = 26^2. ✓

Tip

Tip: A triple only works when you match the right roles --- the largest number is always the hypotenuse. In 88-1515-1717, the legs are 88 and 1515 and the hypotenuse is 1717.

Special Right Triangle: 45-45-90

Concept
The 45-45-90 Triangle
55°

This is an isosceles right triangle: the two legs are equal, and each base angle is 4545^\circ. The hypotenuse is always the leg times 2\sqrt{2}:

leg=x,leg=x,hypotenuse=x2.\text{leg} = x, \qquad \text{leg} = x, \qquad \text{hypotenuse} = x\sqrt{2}.

Going up from a leg, multiply by 2\sqrt{2}. Going down from the hypotenuse, divide by 2\sqrt{2}.

An angle measures a turn.

Example
Legs equal, hypotenuse is leg times 2\sqrt2

Each leg is 55. Then the hypotenuse is 525\sqrt{2} (about 7.077.07). We keep the exact radical 525\sqrt2 as the answer.

Tip

Remember: In a 4545-4545-9090 triangle the hypotenuse is the only side with a 2\sqrt2. Both legs match.

Special Right Triangle: 30-60-90

Concept
The 30-60-90 Triangle
55°

The sides are always in the ratio

short leg=x,long leg=x3,hypotenuse=2x.\text{short leg} = x, \qquad \text{long leg} = x\sqrt{3}, \qquad \text{hypotenuse} = 2x.

The short leg (across from 3030^\circ) is the key. The hypotenuse (across from 9090^\circ) is twice the short leg, and the long leg (across from 6060^\circ) is the short leg times 3\sqrt3.

An angle measures a turn.

Example
Using the xx, x3x\sqrt3, 2x2x pattern

The short leg (opposite 3030^\circ) is 44. Then the hypotenuse is 24=82\cdot 4 = 8 and the long leg is 434\sqrt{3} (about 6.936.93). We keep 434\sqrt3 exact.

Tip

Tip: Always find the short leg first. If you are given the hypotenuse, halve it to get the short leg; then multiply the short leg by 3\sqrt3 for the long leg.

The Three Trig Ratios: SOH-CAH-TOA

Concept
Sine, Cosine, Tangent
θ°adjopphyp

Pick one of the two acute angles and call it θ\theta. Relative to θ\theta, name the sides:

  • [leftmargin=6mm]
  • opposite --- the leg across from θ\theta
  • adjacent --- the leg touching θ\theta (that is not the hypotenuse)
  • hypotenuse --- across from the right angle
sinθ=oppositehypotenuse,cosθ=adjacenthypotenuse,tanθ=oppositeadjacent.\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}.

Remember it as SOH-CAH-TOA.

Opposite, adjacent and hypotenuse are named from the angle.

Example
Reading the three ratios

For angle θ\theta at BB: the opposite leg is 33, the adjacent leg is 44, and the hypotenuse is 55. So

sinθ=35,cosθ=45,tanθ=34.\sin\theta = \tfrac{3}{5}, \qquad \cos\theta = \tfrac{4}{5}, \qquad \tan\theta = \tfrac{3}{4}.
Tip

Remember: “Opposite” and “adjacent” switch when you switch which angle you look at, but the hypotenuse never moves. Always decide which angle is θ\theta first.

Using a Trig Ratio to Find a Missing Side

Concept
Set up, then solve
θ°adjopphyp

When you know one angle and one side and want another side, pick the ratio that uses your two sides (SOH, CAH, or TOA), then solve for the unknown. Your calculator must be in degree mode.

Opposite, adjacent and hypotenuse are named from the angle.

Example
Finding a side with cosine

Find xx, the side adjacent to the 4040^\circ angle, when the hypotenuse is 1010. Adjacent and hypotenuse means cosine (CAH):

cos40=x10    x=10cos4010(0.766)7.7.\cos 40^\circ = \frac{x}{10} \;\Rightarrow\; x = 10\cos 40^\circ \approx 10(0.766) \approx 7.7.

Rounded to the nearest tenth, x7.7x \approx 7.7.

Example
When the unknown is on the bottom

Find the hypotenuse hh when the side opposite 3535^\circ is 66. Opposite and hypotenuse means sine (SOH):

sin35=6h    h=6sin3560.57410.5.\sin 35^\circ = \frac{6}{h} \;\Rightarrow\; h = \frac{6}{\sin 35^\circ} \approx \frac{6}{0.574} \approx 10.5.

When the unknown sits in the denominator, swap it with the trig value.

Tip

Tip: If the unknown is on top, multiply. If the unknown is on the bottom, divide the known side by the trig value.

Using Inverse Trig to Find a Missing Angle

Concept
The inverse buttons
55°

When you know two sides but want the angle, use the inverse functions sin1\sin^{-1}, cos1\cos^{-1}, tan1\tan^{-1} (often the “22nd” + sin/cos/tan keys):

θ=sin1 ⁣(opphyp),    θ=cos1 ⁣(adjhyp),    θ=tan1 ⁣(oppadj).\theta = \sin^{-1}\!\left(\frac{\text{opp}}{\text{hyp}}\right),\;\; \theta = \cos^{-1}\!\left(\frac{\text{adj}}{\text{hyp}}\right),\;\; \theta = \tan^{-1}\!\left(\frac{\text{opp}}{\text{adj}}\right).

The inverse “undoes” the ratio and gives back the angle in degrees.

An angle measures a turn.

Example
Finding an angle with tangent

The leg opposite θ\theta is 55 and the leg adjacent is 88. Two legs means tangent (TOA):

tanθ=58=0.625    θ=tan1(0.625)32.0.\tan\theta = \frac{5}{8} = 0.625 \;\Rightarrow\; \theta = \tan^{-1}(0.625) \approx 32.0^\circ.

Rounded to the nearest tenth of a degree, θ32.0\theta \approx 32.0^\circ.

Tip

Remember: A plain trig button turns an angle into a ratio; an inverse button (1^{-1}) turns a ratio back into an angle. Choose the ratio SOH-CAH-TOA from the two sides you have.

Angles of Elevation & Depression

Concept
Looking up and looking down
55°

An angle of elevation is measured up from the horizontal to an object above you. An angle of depression is measured down from the horizontal to an object below you. Because the two horizontal lines are parallel, the angle of elevation from the ground equals the angle of depression from above (alternate interior angles).

An angle measures a turn.

Example
Height from an angle of elevation

From 5050 ft away, the angle of elevation to the top of a tree is 3232^\circ. Find the height hh. Opposite and adjacent means tangent (TOA):

tan32=h50    h=50tan3250(0.625)31.2 ft.\tan 32^\circ = \frac{h}{50} \;\Rightarrow\; h = 50\tan 32^\circ \approx 50(0.625) \approx 31.2 \text{ ft}.

Rounded to the nearest tenth, the tree is about 31.231.2 ft tall.

Example
An angle of depression

A lifeguard 1010 ft up sees a swimmer at an angle of depression of 2020^\circ. The angle of depression equals the angle of elevation from the swimmer, so at the tower base the angle inside the triangle is 2020^\circ. With the height 1010 opposite that angle and the horizontal distance dd adjacent:

tan20=10d    d=10tan20100.36427.5 ft.\tan 20^\circ = \frac{10}{d} \;\Rightarrow\; d = \frac{10}{\tan 20^\circ} \approx \frac{10}{0.364} \approx 27.5 \text{ ft}.
Tip

Tip: Always sketch the right triangle and label horizontal, vertical, and line of sight. The elevation/depression angle sits between the horizontal and the line of sight --- never at the top of a vertical side.

Going Deeper: Advanced Right-Triangle & Trig Ideas

Concept
Where the 45-45-90 ratio comes from
55°

The ratio is not memorized magic --- it follows from the Pythagorean Theorem. Start with an isosceles right triangle whose two equal legs are each xx. The hypotenuse cc obeys

c2=x2+x2=2x2    c=2x2=x2.c^2 = x^2 + x^2 = 2x^2 \;\Rightarrow\; c = \sqrt{2x^2} = x\sqrt{2}.

Because both legs are equal, the two acute angles are equal, and since they must sum to 9090^\circ each is 4545^\circ. That is exactly the pattern xx, xx, x2x\sqrt2.

An angle measures a turn.

Concept
Where the 30-60-90 ratio comes from

Take an equilateral triangle with every side 2x2x and every angle 6060^\circ. Drop an altitude from the top vertex: it splits the base into two halves of length xx and cuts the top angle into two 3030^\circ pieces, making a right triangle. The short leg (opposite 3030^\circ) is xx and the hypotenuse is 2x2x, so the altitude (the long leg, opposite 6060^\circ) is

h2=(2x)2x2=4x2x2=3x2    h=x3.h^2 = (2x)^2 - x^2 = 4x^2 - x^2 = 3x^2 \;\Rightarrow\; h = x\sqrt{3}.

That gives the pattern xx, x3x\sqrt3, 2x2x.

Concept
Preview: the identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1

In a right triangle with hypotenuse cc, opposite side aa, and adjacent side bb, we have sinθ=a/c\sin\theta = a/c and cosθ=b/c\cos\theta = b/c. Squaring and adding:

sin2θ+cos2θ=a2c2+b2c2=a2+b2c2=c2c2=1,\sin^2\theta + \cos^2\theta = \frac{a^2}{c^2} + \frac{b^2}{c^2} = \frac{a^2+b^2}{c^2} = \frac{c^2}{c^2} = 1,

because a2+b2=c2a^2+b^2=c^2. So this famous Pythagorean identity is just the Pythagorean Theorem divided by c2c^2. (Here sin2θ\sin^2\theta means (sinθ)2(\sin\theta)^2.) It holds for every angle, so if you know one of sinθ\sin\theta or cosθ\cos\theta you can find the other.

Reminder — The Pythagorean identity:sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1
Example
Using the identity

Suppose sinθ=35\sin\theta = \tfrac{3}{5} for an acute angle θ\theta. Find cosθ\cos\theta without drawing the triangle:

cos2θ=1sin2θ=1925=1625    cosθ=1625=45.\cos^2\theta = 1 - \sin^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25} \;\Rightarrow\; \cos\theta = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}.

We take the positive root because θ\theta is acute. (This is just the 33-44-55 triangle in disguise.)

Concept
Beyond right triangles: Law of Sines & Law of Cosines

SOH-CAH-TOA only works in right triangles. For any triangle with angles AA, BB, CC opposite sides aa, bb, cc:

Law of Sines:sinAa=sinBb=sinCc,\text{Law of Sines:}\quad \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c},
Law of Cosines:c2=a2+b22abcosC.\text{Law of Cosines:}\quad c^2 = a^2 + b^2 - 2ab\cos C.

The Law of Sines pairs each angle with its opposite side (use it when you know an angle-side pair). The Law of Cosines is the Pythagorean Theorem with a correction term 2abcosC-2ab\cos C; when C=90C = 90^\circ, cosC=0\cos C = 0 and it collapses back to c2=a2+b2c^2 = a^2+b^2.

Reminder — Law of Sines:asinA=bsinB=csinC\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
Reminder — Law of Cosines:c2=a2+b22abcosCc^{2}=a^{2}+b^{2}-2ab\cos C
Example
Law of Cosines for a non-right triangle

A triangle has sides a=7a = 7 and b=10b = 10 with the included angle C=40C = 40^\circ between them. Find the third side cc:

c2=72+1022(7)(10)cos40=49+100140(0.766)149107.2=41.8.c^2 = 7^2 + 10^2 - 2(7)(10)\cos 40^\circ = 49 + 100 - 140(0.766) \approx 149 - 107.2 = 41.8.
c=41.86.5.c = \sqrt{41.8} \approx 6.5.

Rounded to the nearest tenth, c6.5c \approx 6.5.

Concept
Right triangles in 3D: the space diagonal

A rectangular box with length \ell, width ww, and height hh has a space diagonal dd running corner to opposite corner. Use the Pythagorean Theorem twice: first the diagonal of the base is 2+w2\sqrt{\ell^2 + w^2}, then that base diagonal and the height hh form a second right triangle:

d=2+w2+h2.d = \sqrt{\ell^2 + w^2 + h^2}.

The one clean formula just adds the squares of all three dimensions.

Example
Space diagonal of a box

Find the space diagonal of a 3×4×123 \times 4 \times 12 box.

d=32+42+122=9+16+144=169=13.d = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13.

Notice the base diagonal is 32+42=5\sqrt{3^2+4^2}=5, and then 52+122=13\sqrt{5^2+12^2}=13 --- two nested triples (33-44-55 and 55-1212-1313).

Example
Two observers, one balloon

Two people stand 200200 ft apart on level ground, a balloon directly between them on the line joining them. One measures an angle of elevation of 5050^\circ, the other 4040^\circ. Find the height hh. Let the foot of the balloon be xx ft from the first person, so it is 200x200 - x ft from the second. Each person sees the same height:

h=xtan50,h=(200x)tan40.h = x\tan 50^\circ, \qquad h = (200 - x)\tan 40^\circ.

Set them equal: xtan50=(200x)tan40x\tan 50^\circ = (200 - x)\tan 40^\circ, so 1.19x=(200x)(0.839)1.19x = (200-x)(0.839), giving 1.19x=167.80.839x1.19x = 167.8 - 0.839x, hence 2.03x=167.82.03x = 167.8 and x82.7x \approx 82.7. Then

h=82.7tan5082.7(1.19)98.5 ft.h = 82.7\tan 50^\circ \approx 82.7(1.19) \approx 98.5 \text{ ft}.

Rounded to the nearest tenth, the balloon is about 98.598.5 ft high.

Concept
The altitude on the hypotenuse: geometric means

Drop the altitude from the right angle to the hypotenuse. It splits the hypotenuse into two pieces pp and qq and creates two smaller triangles, each similar to the original. Similarity gives three “geometric mean” relations:

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

The altitude is the geometric mean of the two hypotenuse pieces; each leg is the geometric mean of the whole hypotenuse and the piece next to it.

Example
Finding the altitude by geometric mean

The altitude to the hypotenuse splits it into pieces p=4p = 4 and q=9q = 9. Find the altitude aa:

a=pq=49=36=6.a = \sqrt{p\,q} = \sqrt{4 \cdot 9} = \sqrt{36} = 6.

As a check, the shorter leg is 4(4+9)=52=213\sqrt{4(4+9)} = \sqrt{52} = 2\sqrt{13} and the longer leg is 9(13)=117=313\sqrt{9(13)} = \sqrt{117} = 3\sqrt{13}; indeed (213)2+(313)2=52+117=169=132(2\sqrt{13})^2 + (3\sqrt{13})^2 = 52 + 117 = 169 = 13^2, matching the full hypotenuse 4+9=134+9=13. ✓

Tip

Big picture: Almost every idea here is the Pythagorean Theorem wearing a costume --- the special-triangle ratios, the identity sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1, the Law of Cosines, and the 3D space diagonal all reduce to “add the squares.” When a problem looks new, hunt for the right triangle hiding inside it.

Formulas, Proofs & Tips

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

What it means. In a right triangle the squares on the legs add to the square on the hypotenuse.

Example. Legs 33 and 44: c=32+42=5c=\sqrt{3^2+4^2}=5.

Why it works. Take four copies of the triangle and place them inside a square of side a+ba+b, leaving a tilted square of side cc in the middle. The big square's area is (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2; it is also the four triangles 412ab=2ab4\cdot\tfrac12 ab=2ab plus c2c^2. Cancelling 2ab2ab leaves a2+b2=c2a^2+b^2=c^2.

Tip. cc must be the hypotenuse — the side opposite the right angle, always the longest. The converse also holds: if a2+b2=c2a^2+b^2=c^2 the triangle is right-angled.

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 2212=3\sqrt{2^2-1^2}=\sqrt3.

Tip. The short leg is always opposite the 3030^\circ angle. Match sides to angles before assigning 11, 3\sqrt3, 22.

Tip
The trigonometric ratios (SOH-CAH-TOA)
sinθ=opphyp,cosθ=adjhyp,tanθ=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. tanθ=sinθcosθ\tan\theta=\tfrac{\sin\theta}{\cos\theta}, since dividing opp/hyp by adj/hyp cancels the hypotenuse.