Right Triangle Trigonometry: Ratios, Missing Sides, and Angles

Learning Targets

By the end of this session, students should be able to achieve the following objectives:

  • Define and identify the six trigonometric ratios.

  • Evaluate trigonometric ratios based on 주어진 right triangles.

  • Solve for the missing sides and angles of a right triangle.

Review: Right Triangles and the Pythagorean Theorem

  • Definition of a Right Triangle: A triangle in which one angle is exactly a right angle (90∘90^{\circ}).

  • The Pythagorean Theorem: In a right triangle with a hypotenuse of length cc and legs of lengths aa and bb, the following relationship holds:

a2+b2=c2a^2 + b^2 = c^2

  • Practical Application Examples:

    • Example 1: Find the missing side cc if a=4a = 4 and b=3b = 3.

      • Solution: 42+32=c24^2 + 3^2 = c^2.

      • 16+9=c216 + 9 = c^2

      • 25=c225 = c^2

      • c=25=5c = \sqrt{25} = 5

      • Note: Only the positive square root is taken because cc represents distance, which is always positive.

    • Example 2: Find side aa if b=4b = 4 and c=8c = 8.

      • Solution: a2+42=82a^2 + 4^2 = 8^2

      • a2+16=64a^2 + 16 = 64

      • a2=64−16=48a^2 = 64 - 16 = 48

      • a=48=16×3=43a = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}

    • Example 3: Find cc if a=5a = 5 and b=12b = 12.

      • Result: c=13c = 13

    • Example 4: Find bb if a=6a = 6 and c=9c = 9.

      • Result: b=45=35b = \sqrt{45} = 3\sqrt{5}

The Six Trigonometric Ratios

If θ\theta is an acute angle (0^{\circ} < \theta < 90^{\circ}) in a right triangle, six ratios can be formed that depend only on the angle θ\theta.

Primary Trigonometric Ratios (SOH-CAH-TOA)

  • Sine (sin): The ratio of the length of the opposite leg to the length of the hypotenuse.

sin⁡(θ)=Opposite legHypotenuse=OH\sin(\theta) = \frac{\text{Opposite leg}}{\text{Hypotenuse}} = \frac{O}{H}

  • Cosine (cos): The ratio of the length of the adjacent leg to the length of the hypotenuse.

cos⁡(θ)=Adjacent legHypotenuse=AH\cos(\theta) = \frac{\text{Adjacent leg}}{\text{Hypotenuse}} = \frac{A}{H}

  • Tangent (tan): The ratio of the length of the opposite leg to the length of the adjacent leg.

tan⁡(θ)=Opposite legAdjacent leg=OA\tan(\theta) = \frac{\text{Opposite leg}}{\text{Adjacent leg}} = \frac{O}{A}

Reciprocal Trigonometric Ratios

  • Cosecant (csc): The reciprocal of sine.

csc⁡(θ)=HypotenuseOpposite leg=HO\csc(\theta) = \frac{\text{Hypotenuse}}{\text{Opposite leg}} = \frac{H}{O}

  • Secant (sec): The reciprocal of cosine.

sec⁡(θ)=HypotenuseAdjacent leg=HA\sec(\theta) = \frac{\text{Hypotenuse}}{\text{Adjacent leg}} = \frac{H}{A}

  • Cotangent (cot): The reciprocal of tangent.

cot⁡(θ)=Adjacent legOpposite leg=AO\cot(\theta) = \frac{\text{Adjacent leg}}{\text{Opposite leg}} = \frac{A}{O}

Evaluating Trigonometric Ratios: Worked Examples

Example 1: Triangle with sides 12 and 13

Given a right triangle ABCABC where leg b=12b = 12 and hypotenuse c=13c = 13.

  1. Find the third side (a):

    • a2+122=132a^2 + 12^2 = 13^2

    • a2+144=169a^2 + 144 = 169

    • a2=25→a=5a^2 = 25 \rightarrow a = 5

  2. Evaluate Ratios:

    • sin⁡(A)=OH=513\sin(A) = \frac{O}{H} = \frac{5}{13}

    • cos⁡(B)=AH=513\cos(B) = \frac{A}{H} = \frac{5}{13}

    • tan⁡(A)=OA=512\tan(A) = \frac{O}{A} = \frac{5}{12}

    • csc⁡(B)=HO=1312\csc(B) = \frac{H}{O} = \frac{13}{12}

    • sec⁡(A)=HA=1312\sec(A) = \frac{H}{A} = \frac{13}{12}

    • cot⁡(B)=AO=512\cot(B) = \frac{A}{O} = \frac{5}{12}

Practice Exercise 1: Triangle with sides 8 and 10

Given leg XZ=8XZ = 8 and hypotenuse XY=10XY = 10 for triangle XYZXYZ. The third side (found via Pythagorean theorem) is 66.

  • sin⁡(X)=610=35\sin(X) = \frac{6}{10} = \frac{3}{5}

  • tan⁡(Z)=86=43\tan(Z) = \frac{8}{6} = \frac{4}{3}

  • csc⁡(X)=106=53\csc(X) = \frac{10}{6} = \frac{5}{3}

  • cot⁡(Z)=68=34\cot(Z) = \frac{6}{8} = \frac{3}{4}

Example 2: Triangle with rationalized values

Given leg TU=8TU = 8 and leg TV=4TV = 4. Determine the ratios for angle VV and UU.

  1. Find Hypotenuse (c):

    • 82+42=c28^2 + 4^2 = c^2

    • 64+16=8064 + 16 = 80

    • c=80=45c = \sqrt{80} = 4\sqrt{5}

  2. Evaluate Ratios (including rationalizing the denominator):

    • sin⁡(V)=845=25×55=255\sin(V) = \frac{8}{4\sqrt{5}} = \frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5}

    • cos⁡(V)=445=15×55=55\cos(V) = \frac{4}{4\sqrt{5}} = \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}

    • tan⁡(U)=48=12\tan(U) = \frac{4}{8} = \frac{1}{2}

    • csc⁡(U)=454=5\csc(U) = \frac{4\sqrt{5}}{4} = \sqrt{5}

    • sec⁡(U)=458=52\sec(U) = \frac{4\sqrt{5}}{8} = \frac{\sqrt{5}}{2}

    • cot⁡(V)=48=12\cot(V) = \frac{4}{8} = \frac{1}{2}

Finding the Missing Side of a Right Triangle

When given one side and one acute angle, you can find the other sides using trigonometric ratios.

  • Example 3: Triangle with Hypotenuse = 1515 and angle θ=20∘\theta = 20^{\circ}.

    • Find Opposite (VT): Use Sine.

      • sin⁡(20∘)=VT15\sin(20^{\circ}) = \frac{VT}{15}

      • VT=15×sin⁡(20∘)≈5.13VT = 15 \times \sin(20^{\circ}) \approx 5.13

    • Find Adjacent (UT): Use Cosine.

      • cos⁡(20∘)=UT15\cos(20^{\circ}) = \frac{UT}{15}

      • UT=15×cos⁡(20∘)≈14.10UT = 15 \times \cos(20^{\circ}) \approx 14.10

  • Practice Exercise 3:

    • Problem 1: Hypotenuse = 1313, angle = 65∘65^{\circ}.

      • AC=13×cos⁡(65∘)≈5.49AC = 13 \times \cos(65^{\circ}) \approx 5.49

      • BC=13×sin⁡(65∘)≈11.78BC = 13 \times \sin(65^{\circ}) \approx 11.78

    • Problem 2: Opposite = 12.512.5, angle = 38∘38^{\circ}.

      • Find Adjacent (BC): Use Tangent.

        • tan⁡(38∘)=12.5BC\tan(38^{\circ}) = \frac{12.5}{BC}

        • Using the Extremes-Switch Law of Proportion: BC=12.5tan⁡(38∘)≈16.0BC = \frac{12.5}{\tan(38^{\circ})} \approx 16.0

      • Find Hypotenuse (AB): Use Sine.

        • sin⁡(38∘)=12.5AB\sin(38^{\circ}) = \frac{12.5}{AB}

        • AB=12.5sin⁡(38∘)≈20.30AB = \frac{12.5}{\sin(38^{\circ})} \approx 20.30

Finding an Acute Angle of a Right Triangle

To find an angle when sides are known, use inverse trigonometric functions (sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1}, tan⁡−1\tan^{-1}).

  • Formulas:

    • If sin⁡(A)=ac\sin(A) = \frac{a}{c}, then A=sin⁡−1(ac)A = \sin^{-1}\left(\frac{a}{c}\right).

    • If cos⁡(A)=bc\cos(A) = \frac{b}{c}, then A=cos⁡−1(bc)A = \cos^{-1}\left(\frac{b}{c}\right).

    • If tan⁡(A)=ab\tan(A) = \frac{a}{b}, then A=tan⁡−1(ab)A = \tan^{-1}\left(\frac{a}{b}\right).

  • Example 4: Triangle with leg TU=4TU = 4 and hypotenuse UV=12UV = 12.

    • Find angle U: Angle U is adjacent to the side of 4.

      • sin⁡(V)=412\sin(V) = \frac{4}{12}

      • U=sin⁡−1(412)≈19.47∘U = \sin^{-1}\left(\frac{4}{12}\right) \approx 19.47^{\circ}

    • Find angle V:

      • Method 1: V=90∘−19.47∘=70.53∘V = 90^{\circ} - 19.47^{\circ} = 70.53^{\circ}.

      • Method 2 (using Cosine): cos⁡(V)=412→V=cos⁡−1(412)≈70.53∘\cos(V) = \frac{4}{12} \rightarrow V = \cos^{-1}\left(\frac{4}{12}\right) \approx 70.53^{\circ}.

  • Practice Exercise 4: Determine angles in a triangle with legs 1111 and 1212.

    • Find angle A: Opposite is 11, Adjacent is 12.

      • tan⁡(A)=1112→A=tan⁡−1(1112)≈42.51∘\tan(A) = \frac{11}{12} \rightarrow A = \tan^{-1}\left(\frac{11}{12}\right) \approx 42.51^{\circ}

    • Find angle N: Opposite is 12, Adjacent is 11.

      • tan⁡(N)=1211→N=tan⁡−1(1211)≈47.49∘\tan(N) = \frac{12}{11} \rightarrow N = \tan^{-1}\left(\frac{12}{11}\right) \approx 47.49^{\circ}

Extended Practice

Section A: Trigonometric Functions

  1. Evaluate the six trigonometric functions for the angle in the provided figures (Triangle 1: legs 3, 4; Triangle 2: specified angle).

Section B: Solve for x

  1. In a triangle with angle 37∘37^{\circ} and hypotenuse 1111, solve for leg xx.

  2. In a triangle with angle 60∘60^{\circ} and leg 1111, solve for hypotenuse xx.

Section C: Solve for Angle θ\theta

  1. Given hypotenuse 1313 and opposite leg 44, find θ\theta.

  2. Given opposite leg 22 and adjacent leg 1313, find θ\theta.