Right-Triangle Trigonometry & Special Triangles

Fundamental Right-Triangle Concepts

  • A right triangle contains one 9090^\circ angle.

  • Side names relative to an acute angle θ\theta:

    • Hypotenuse (c): side opposite the right angle; always the longest.

    • Opposite (o): side opposite the chosen angle θ\theta.

    • Adjacent (a): side next to θ\theta that is not the hypotenuse.

  • Classic theorem:
    a2+b2=c2a^2 + b^2 = c^2 (Pythagorean Theorem).

  • Pythagorean triples (integer solutions): e.g. 3453-4-5, 512135-12-13, 68106-8-10.

    • Historical note: Pythagoreans zealously guarded discoveries of new triples ("first mathematical cult").

Finding Missing Sides (examples)

  • Example 1: sides 55 and 1212, find hypotenuse.

    • 25+144=c2c=1325 + 144 = c^2 \Rightarrow c = 13.

  • Example 2: sides 1010 (hyp) and 44 (leg), find other leg.

    • a2+42=102a=84=221a^2 + 4^2 = 10^2 \Rightarrow a = \sqrt{84}=2\sqrt{21}.

  • Example 3: legs 2,32,3c=13c = \sqrt{13} (simplify radicals only; do not decimal-approximate in exact answers).

Right-Triangle Definitions of Trig Ratios

  • Using θ\theta in a right triangle:

    • sinθ=OppositeHypotenuse\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}

    • cosθ=AdjacentHypotenuse\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}

    • tanθ=OppositeAdjacent\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}}

    • Reciprocals:

    • cscθ=1sinθ=HypotenuseOpposite\csc\theta = \dfrac{1}{\sin\theta}=\dfrac{\text{Hypotenuse}}{\text{Opposite}}

    • secθ=1cosθ=HypotenuseAdjacent\sec\theta = \dfrac{1}{\cos\theta}=\dfrac{\text{Hypotenuse}}{\text{Adjacent}}

    • cotθ=1tanθ=AdjacentOpposite\cot\theta = \dfrac{1}{\tan\theta}=\dfrac{\text{Adjacent}}{\text{Opposite}}

  • Keep two "tattooed" reference systems: one arm = unit-circle definitions, other arm = right-triangle definitions.

Unit-Circle vs. Right-Triangle View

  • Unit circle (radius 11) point (x,y)(x,y) with same θ\theta:

    • sinθ=y\sin\theta = y, cosθ=x\cos\theta = x, tanθ=yx\tan\theta = \dfrac{y}{x}.

  • Build a right triangle by dropping a perpendicular → leg lengths xx and yy, hypotenuse 11.

    • Matches right-triangle ratios exactly; thus both definitions are equivalent.

Reciprocal & Quotient Identities (review)

  • Reciprocal pairs all satisfy f(θ)f1(θ)=1f(\theta)\cdot f^{-1}(\theta)=1:

    • sinθcscθ=1\sin\theta\,\csc\theta = 1, cosθsecθ=1\cos\theta\,\sec\theta = 1, tanθcotθ=1\tan\theta\,\cot\theta = 1.

  • Quotient:

    • tanθ=sinθcosθ\tan\theta = \dfrac{\sin\theta}{\cos\theta}, cotθ=cosθsinθ\cot\theta = \dfrac{\cos\theta}{\sin\theta}.

Pythagorean Identities (derived from a2+b2=c2a^2+b^2=c^2)

  • sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1

  • 1+tan2θ=sec2θ1 + \tan^2\theta = \sec^2\theta

  • 1+cot2θ=csc2θ1 + \cot^2\theta = \csc^2\theta

Worked Example: Finding Remaining Ratios from Two

Given sinθ=23,  cosθ=53\sin\theta = \dfrac{2}{3},\; \cos\theta = \dfrac{\sqrt{5}}{3}.

  • Reciprocal quickly: cscθ=32,  secθ=355\csc\theta=\dfrac{3}{2},\; \sec\theta=\dfrac{3\sqrt{5}}{5}.

  • tanθ=sincos=2/35/3=25=255\tan\theta = \dfrac{\sin}{\cos}=\dfrac{2/3}{\sqrt{5}/3}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}.

  • cotθ=1tanθ=52\cot\theta = \dfrac{1}{\tan\theta}=\dfrac{\sqrt{5}}{2}.

Special Right Triangles

30-60-90 Triangle

  • Start with equilateral \triangle (all sides 22). Bisect to form:

    • Hypotenuse =2=2

    • Short leg (opposite 3030^\circ) =1=1

    • Long leg (opposite 6060^\circ) =3=\sqrt{3}

  • Key ratios:

    • sin30=12\sin30^\circ = \tfrac12

    • cos30=32\cos30^\circ = \dfrac{\sqrt{3}}{2}

    • tan30=13=33\tan30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}

    • sin60=32\sin60^\circ = \dfrac{\sqrt{3}}{2}

    • cos60=12\cos60^\circ = \tfrac12

    • tan60=3\tan60^\circ = \sqrt{3}

  • For any 30-60-90 triangle: hypotenuse =2k=2k, short leg =k=k, long leg =k3=k\sqrt{3}.

45-45-90 Triangle

  • Isosceles right triangle; let legs =a=a.

    • a2+a2=22a=2a^2+a^2=2^2\Rightarrow a=\sqrt{2} when hypotenuse set to 22.

  • Ratios (same for both acute angles):

    • sin45=cos45=22\sin45^\circ = \cos45^\circ = \dfrac{\sqrt{2}}{2}

    • tan45=1\tan45^\circ = 1

    • csc45=sec45=2\csc45^\circ = \sec45^\circ = \sqrt{2}

    • cot45=1\cot45^\circ = 1

Rationalizing Denominators

  • If a radical appears in a denominator, multiply by an appropriate 11 (e.g. nn\dfrac{\sqrt{n}}{\sqrt{n}}) to move the radical to the numerator.

    • Example: 15×55=55\dfrac{1}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}.

Calculator Usage Tips

  • Always set MODE correctly:

    • For triangles, switch from default Radians to Degrees.

  • Keys: SIN, COS, TAN expect degree input when in degree mode.

  • Calculators usually output decimal approximations. Recognize common special-angle decimals:

    • sin30=0.5\sin30^\circ=0.5

    • cos300.8660254(3/2)\cos30^\circ\approx0.8660254\ldots (\sqrt{3}/2)

    • cos450.7071067(2/2)\cos45^\circ\approx0.7071067\ldots (\sqrt{2}/2)

  • Double-check odd answers by verifying largest side ↔ largest angle, etc.

Solving Right Triangles (General Procedure)

  1. Sketch and label: right angle at CC, other angles A,BA,B; opposite sides a,ba,b; hypotenuse cc.

  2. Easiest first: find missing acute angle using A+B=90A+B=90^\circ.

  3. Prefer given data: when possible use sides/angles provided in the problem instead of values you computed (reduces rounding errors).

  4. Choose ratio matching the unknown:

    • Need hypotenuse? Probably use sin\sin or cos\cos.

    • Need opposite or adjacent? Pick among sin,cos,tan\sin,\cos,\tan accordingly.

  5. Solve algebraically, then compute with calculator (degree mode).

  6. Check: largest angle ↔ largest side; smallest angle ↔ smallest side; a2+b2c2a^2+b^2\approx c^2.

Example Walk-Through 1

Given B=69,  b=37B=69^\circ,\; b=37.

  • A=9069=21A = 90-69 = 21^\circ.

  • Solve for cc using cos\cos (adjacent/hyp): cos69=37/cc=37/cos69103.2\cos69^\circ = 37/c \Rightarrow c = 37/\cos69^\circ \approx 103.2.

  • Solve for aa using tan\tan (opp/adj): tan69=a/37a=37tan69100.4\tan69^\circ = a/37 \Rightarrow a = 37\tan69^\circ \approx 100.4.

Example Walk-Through 2

Given B=57.4,  b=70B=57.4^\circ,\; b=70.

  • A=32.6A = 32.6^\circ.

  • c=70sin57.483.1c = \dfrac{70}{\sin57.4^\circ} \approx 83.1.

  • a=70tan57.444.8a = \dfrac{70}{\tan57.4^\circ} \approx 44.8.

Example Walk-Through 3

Given c=343.6,  B=32.17c = 343.6,\; B = 32.17^\circ.

  • A=57.83A = 57.83^\circ.

  • a=ccosB=343.6cos32.17290.8a = c\cos B = 343.6\cos32.17^\circ \approx 290.8.

  • b=csinB=343.6sin32.17182.9b = c\sin B = 343.6\sin32.17^\circ \approx 182.9.

Cofunction Identities

  • Definition: Trig functions paired with a "co-" version (sine ↔ cosine, tangent ↔ cotangent, secant ↔ cosecant).

  • Rule: Cofunctions of complementary angles are equal.

    • If A+B=90A + B = 90^\circ,

    • sinA=cosB\sin A = \cos B,

    • tanA=cotB\tan A = \cot B,

    • secA=cscB\sec A = \csc B.

  • Example: sin30=cos60=0.5\sin30^\circ = \cos60^\circ = 0.5; sin57.83cos32.17\sin57.83^\circ \approx \cos32.17^\circ.

Key Study Tips

  • Memorize exactly: all six trig ratios, reciprocal pairs, quotient rules.

  • Know by heart values for 30,45,6030^\circ, 45^\circ, 60^\circ (both sine & cosine—and thus tangent).

  • Practice rationalizing denominators quickly.

  • Check calculator mode before every computation.

  • Label diagrams consistently; avoid mixing derived numbers if an original value is available.

  • For exam #2 (not trig): focus review on quadratic → logarithmic functions; trig resumes afterward.