Right Triangles and Trigonometry: Chapter Review

Vocabulary and notation

Hypotenuse and legs

The hypotenuse lies opposite a right triangle's right angle and is its longest side. The other two sides are legs.

Hypotenuse altitude and foot

The perpendicular from the right-angle vertex to the hypotenuse meets its open segment. Its meeting point is the foot; it need not be the midpoint.

Altitude notation

In $ABC$ right at $C$, let $CH\perp AB$, with $H$ interior. Lengths are $a=BC$, $b=CA$, $c=AB$, $p=AH$, $q=HB$, and $h=CH$.

Geometric mean

For positive numbers $u,v$, the geometric mean is $\sqrt{uv}$, the positive value between them in the equal ratios $u/\sqrt{uv}=\sqrt{uv}/v$.

Pythagorean triple

Three positive integers whose two smaller squares add to the largest square. A common positive length factor preserves the right-triangle relationship.

Reference angle, opposite, and adjacent

Select an acute angle first. Its opposite leg is across from it; its adjacent leg meets it and the right angle. The hypotenuse is excluded from adjacent leg.

Inverse trigonometric operation

An operation taking a valid side ratio and returning the acute angle that produces it. The notation $\sin^{-1}$ means inverse sine, not a reciprocal.

Elevation and depression

An elevation angle turns upward from the horizontal through the eye; a depression angle turns downward from it. A vertical reference gives the complementary angle.

Rise, run, and sloped length

Rise is vertical change; run is horizontal distance. The sloped length is the hypotenuse of the resulting right triangle. Percent gradient compares rise with run.

Formulas and theorems

• Hypotenuse altitude relationships

  $$\begin{gathered} \triangle ABC\sim\triangle ACH\\ \triangle ACH\sim\triangle CBH\\ c=p+q\\ h^2=pq\\ b^2=cp\\ a^2=cq \end{gathered}$$

  Text description

  Triangle ABC has its right angle at C above hypotenuse AB. The perpendicular CH meets AB at interior point H. Length p is AH, q is HB, and h is CH. Leg b is CA and meets piece p at A. Leg a is BC and meets piece q at B. A bracket labels the whole hypotenuse c, distinguishing it from either piece.

  Use when Use the altitude notation for a nondegenerate triangle right at $C$. AA establishes the ordered similarities. Each leg uses the whole hypotenuse and the piece at its own endpoint.

  e.g. With hypotenuse pieces $AH=5$ cm and $HB=14$ cm, find the altitude: $CH=\sqrt{70}$ cm.

• Pythagorean theorem

  $$a^2+b^2=c^2$$

  Use when A nondegenerate right triangle has legs $a,b$ and hypotenuse $c$, all in common units. A missing leg uses subtraction of squares; every recovered length is the positive root.

  e.g. A right triangle has legs $9$ cm and $14$ cm; find the exact hypotenuse: $\sqrt{277}$ cm.

• Converse and squared-side classification

  $$\begin{gathered} z^2<x^2+y^2:\quad\text{acute}\\ z^2=x^2+y^2:\quad\text{right}\\ z^2>x^2+y^2:\quad\text{obtuse} \end{gathered}$$

  Use when The positive lengths form a nondegenerate triangle, $z$ is longest, and $x+y>z$. Equality identifies the right angle opposite $z$; inequalities classify the largest angle.

  e.g. Classify lengths $11,14,19$: they close, and $19^2>11^2+14^2$, so the triangle is obtuse.

• Common triples and scaling

  $$\begin{gathered} (3,4,5),\ (5,12,13)\\ (7,24,25),\ (8,15,17)\\ (ka)^2+(kb)^2=(kc)^2 \end{gathered}$$

  Use when The original lengths satisfy $a^2+b^2=c^2$ and the same $k>0$ multiplies every length. Noninteger factors still preserve rightness; the resulting lengths need not be integers.

• 45-45-90 lengths

  $$\begin{gathered} \text{legs}:\ s,s\\ \text{hypotenuse}:\ s\sqrt2 \end{gathered}$$

  Use when The right triangle has equal legs, equivalently two $45^\circ$ angles, and $s>0$.

  e.g. A 45-45-90 triangle has hypotenuse $14\sqrt3$ cm; find either leg: $7\sqrt6$ cm.

• 30-60-90 lengths

  $$\begin{gathered} \text{opposite }30^\circ:\ s\\ \text{opposite }60^\circ:\ s\sqrt3\\ \text{hypotenuse}:\ 2s \end{gathered}$$

  Use when The acute angles are $30^\circ$ and $60^\circ$ and the short leg is $s>0$. Identify roles by opposite angles, regardless of orientation.

  e.g. A 30-60-90 triangle has long leg $11\sqrt6$ m; find its short leg: $11\sqrt2$ m.

• Square diagonal and equilateral altitude

  $$\begin{gathered} d_{\text{square}}=s\sqrt2\\ h_{\text{equilateral}}=\frac{e\sqrt3}{2} \end{gathered}$$

  Use when The square side is $s>0$; the equilateral side is $e>0$. An equilateral vertex-to-midpoint segment forms perpendicular congruent halves by SSS.

  e.g. An equilateral triangle has side $26$ mm; find its exact altitude: $13\sqrt3$ mm.

• Acute-angle side ratios

  $$\begin{gathered} \sin\theta=\frac ac\\ \cos\theta=\frac bc\\ \tan\theta=\frac ab \end{gathered}$$

  Text description

  The selected acute angle theta is at A at the lower left. C is the lower right vertex and has a right-angle mark. B is above C. Vertical leg BC, named a, is opposite theta. Horizontal leg CA, named b, is adjacent to theta. Sloping side AB, named c, is the hypotenuse. Selecting B would exchange the leg roles but leave the hypotenuse unchanged.

  Use when Triangle $ABC$ is right at $C$, with $a=BC$, $b=CA$, $c=AB$, and $\theta$ at $A$. All angles use degrees. AA makes these ratios independent of size. Sine and cosine lie strictly between $0$ and $1$; tangent is positive.

• Complementary acute angles

  $$\begin{gathered} \phi=90^\circ-\theta\\ \sin\phi=\cos\theta\\ \cos\phi=\sin\theta\\ \tan\phi=\frac1{\tan\theta} \end{gathered}$$

  Use when The angles $\theta,\phi$ are acute and complementary. Switching the reference angle exchanges the two legs.

  e.g. Given complementary acute angles with $\tan\theta=5/8$, find $\tan\phi$: $8/5$.

• Acute-angle identities

  $$\begin{gathered} \tan\theta=\frac{\sin\theta}{\cos\theta}\\ \sin^2\theta+\cos^2\theta=1 \end{gathered}$$

  Use when $0^\circ<\theta<90^\circ$, so cosine is positive and division is valid. The notation $\sin^2\theta$ means $(\sin\theta)^2$.

  e.g. For acute $\theta$ with $\sin\theta=5/6$, find the cosine: $\cos\theta=\sqrt{11}/6$.

• Exact special-angle values

  $$\begin{gathered} \sin30^\circ=\cos60^\circ=\frac12\\ \cos30^\circ=\sin60^\circ=\frac{\sqrt3}{2}\\ \sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2}\\ \tan30^\circ=\frac{\sqrt3}{3}\\ \tan45^\circ=1\\ \tan60^\circ=\sqrt3 \end{gathered}$$

  Use when These exact ratios come from the 30-60-90 and 45-45-90 side relationships; angle measures are in degrees.

  e.g. Evaluate $\tan60^\circ\cos30^\circ$ exactly: $3/2$.

• Recover an acute angle

  $$\begin{gathered} \theta=\sin^{-1}\!\left(\frac{\text{opp}}{\text{hyp}}\right)\\ \theta=\cos^{-1}\!\left(\frac{\text{adj}}{\text{hyp}}\right)\\ \theta=\tan^{-1}\!\left(\frac{\text{opp}}{\text{adj}}\right) \end{gathered}$$

  Use when Choose the applicable equation in DEGREE mode. Here $\text{opp}$, $\text{adj}$, and $\text{hyp}$ are the opposite leg, adjacent leg, and hypotenuse lengths. Acute sine/cosine inverse inputs satisfy $0<r<1$; tangent inverse inputs satisfy $r>0$.

  e.g. An acute angle has cosine $0.72$; find it to the nearest tenth: $\theta\approx43.9^\circ$.

• Height from an elevation observation

  $$\begin{gathered} \tan\theta=\frac{H-e}{d}\\ H=e+d\tan\theta \end{gathered}$$

  Text description

  The observer eye E lies a distance e above level ground. A horizontal segment from the eye meets the vertical object above its base, and a sloping sight line reaches the top. The acute elevation theta is at the eye. The right triangle's horizontal run is d and its vertical rise is H minus e. A separate full-height bracket spans the object from ground to top, showing why e must be included in the whole height H.

  Use when A vertical object and observer share level ground and a vertical sight plane. Here $d>0$ is horizontal run, $e$ is eye height above base level, $H$ is object height, and $\theta$ is acute elevation from horizontal. Adjust $e$ for any known ground-level difference.

  e.g. On level ground in one vertical sight plane, an observer's eye is $1.90$ m high and a vertical pole is $16.8$ m tall. With top elevation $27^\circ$, find the horizontal run to the nearest tenth: $d\approx29.2$ m.

• Ramp and gradient relationships

  $$\begin{gathered} \tan\theta=\frac rd\\ \sin\theta=\frac rL\\ \cos\theta=\frac dL\\ L^2=d^2+r^2\\ \text{gradient}=100\left(\frac rd\right)\% \end{gathered}$$

  Use when A straight ramp has positive vertical rise $r$, horizontal run $d$, surface length $L$, and acute inclination $\theta$ measured from horizontal. Use common length units.

  e.g. A ramp rises $0.90$ m over horizontal run $7.20$ m; find its gradient: $12.5\%$.

• Same-side survey stations

  $$\begin{gathered} u=x\tan\beta\\ u=(x+s)\tan\alpha\\ H=e+u \end{gathered}$$

  Use when Far station, near station, and vertical target base occur in that order on one horizontal line. Station separation is $s>0$, near run is $x>0$, equal eye height is $e$, and both sight the same top above them in one vertical plane. Far elevation $\alpha$ is smaller than near elevation $\beta$.

  e.g. With separation $16$ m, far elevation $30^\circ$, and near elevation $60^\circ$, find the near horizontal run: $x=8$ m.

Problem types, step by step

Exam traps

• Trap Matching vertices by their position in a drawing

  Fix Match right and acute angles in order. A shared vertex need not correspond to itself in the altitude triangles.

• Trap Assuming every hypotenuse altitude is a median

  Fix Perpendicularity gives right angles, not equal pieces. Keep $p$ and $q$ distinct unless equality is established.

• Trap Confusing whole, piece, square, and length

  Fix Use $c=p+q$. The altitude uses $pq$; the leg at $A$ uses $cp$. Take a positive square root to recover a length.

• Trap Applying Pythagoras without identifying a right triangle

  Fix Locate a stated or proved right angle. A label $c$ alone proves nothing; for a missing leg, subtract squares rather than adding.

• Trap Classifying lengths that cannot close

  Fix Check the strict triangle inequality first. A flat equality case has no acute, right, or obtuse classification.

• Trap Halving every special triangle's hypotenuse

  Fix Halving gives the short leg only in a 30-60-90 triangle. For equal-leg right triangles, divide by $\sqrt2$.

• Trap Calling the hypotenuse an adjacent leg

  Fix Adjacent leg excludes the hypotenuse. Switching the acute reference angle exchanges the legs while keeping the hypotenuse.

• Trap Dropping the squares in the Pythagorean identity

  Fix The squared sine and cosine values add to $1$. Their unsquared sum need not. Also, $\sqrt{u+v}$ is not generally $\sqrt u+\sqrt v$.

• Trap Confusing inverse operations, reciprocals, and inputs

  Fix An inverse returns an angle in DEGREE mode. Acute sine/cosine inputs lie strictly between zero and one; positive tangent inputs may exceed one.

• Trap Rounding early or using approximate data as exact proof

  Fix Keep full intermediate precision and mark rounded outputs with $\approx$. Near agreement of measured side squares does not prove an exactly right angle.

• Trap Using the wrong distance or height reference

  Fix Tangent uses horizontal run and vertical difference. Include eye elevation relative to the target base; a sloping tape measurement alone supplies neither required quantity.

• Trap Treating percent gradient or station separation as an angle rule

  Fix A gradient is a rise/run percentage, not degrees. Station-run sums and differences require the stated collinearity and order; doubling a tangent ratio does not double its angle.