Grade 11 College Math - Primary Trigonometry, Sine Law, and Cosine Law

Review of Primary Trigonometric Ratios

• The Pythagorean Theorem:

  • Applies to any right-angled triangle containing a $90^\circ$ angle.

  • Mathematical formula relating side lengths:     $a^2 + b^2 = c^2$

  • Variables:

  • $a$ and $b$ represent the lengths of the two legs forming the right angle.

  • $c$ represents the length of the hypotenuse.

  • Component definitions:

  • Hypotenuse: The longest side of a right triangle, situated directly opposite the right ($90^\circ$) angle.

  • Legs: The two sides that intersect to form the right ($90^\circ$) angle.

  • Formulas for determining a missing leg length:

  • Missing side $a$:       $a = \sqrt{c^2 - b^2}$

  • Missing side $b$:       $b = \sqrt{c^2 - a^2}$

• Naming Sides Relative to a Reference Angle:

  • Opposite Side: The side directly across from the specified reference angle.

  • Adjacent Side: The side adjacent to the reference angle that is not the hypotenuse.

• Primary Trigonometric Ratios (SOH CAH TOA):

  • Ratios relate the acute angles of a right triangle to its side lengths:     $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$     $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$     $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$

  • Trigonometric values can be represented as fractions or expanded decimals.

• Procedure for Finding Missing Side Lengths or Angles:

  1. Identify the reference angle.

  2. Label the triangle's sides (opposite, adjacent, hypotenuse) based on the reference angle.

  3. Select the appropriate trigonometric ratio using SOH CAH TOA based on given parameters.

  4. Formulate the trigonometric equation and cross-multiply to solve for the unknown value.

• Calculator Evaluation Practice:

  • Evaluation to four decimal places:

  • $\sin(37^\circ) = 0.6018$

  • $\cos(60^\circ) = 0.5000$

  • $\tan(78^\circ) = 4.7046$

  • $\cos(83^\circ) = 0.1219$

• Worked Examples: Side & Angle Ratios:

  • Example 1: Ratios for Reference Angle $\angle C$:

  • Triangle $\Delta ABC$ with legs $AB = 3\,\text{cm}$, $BC = 4\,\text{cm}$, and hypotenuse $AC = 5\,\text{cm}$, right-angled at $B$.

  • Relative to $\angle C$: Opposite = $3\,\text{cm}$, Adjacent = $4\,\text{cm}$, Hypotenuse = $5\,\text{cm}$.

  • Ratios:       $\sin(C) = \frac{3}{5} = 0.6000$       $\cos(C) = \frac{4}{5} = 0.8000$       $\tan(C) = \frac{3}{4} = 0.7500$

  • Example 2: Ratios for Reference Angle $\angle A$:

  • Same triangle $\Delta ABC$.

  • Relative to $\angle A$: Opposite = $4\,\text{cm}$, Adjacent = $3\,\text{cm}$, Hypotenuse = $5\,\text{cm}$.

  • Ratios:       $\sin(A) = \frac{4}{5} = 0.8000$       $\cos(A) = \frac{3}{5} = 0.6000$       $\tan(A) = \frac{4}{3} \approx 1.3333$

  • Example 3: Finding Unknown Side Lengths:

  • Problem A: Find side $a$ in right triangle $\Delta ABC$ where $\angle A = 43^\circ$, hypotenuse $AC = 67\,\text{cm}$, right angle at B$.\n      \sin(43^\circ) = \frac{a}{67}\n      a = 67 \times \sin(43^\circ) \approx 45.7\,\text{cm}\n - **Problem B**: Find side y$in right triangle$\Delta XYZ$where$\angle X = 22^\circ$, adjacent side$XY = 25\,\text{cm}$, right angle at$Y$.       $\cos(22^\circ) = \frac{25}{y}$       $y = \frac{25}{\cos(22^\circ)} \approx 27.0\,\text{cm}$

Inverse Trigonometric Functions and Angles

• Inverse Trigonometric Functions:

  • Used to find an unknown angle when two side lengths are given.

  • Notations: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$.

• Calculator Evaluation for Angles (Nearest Degree):

  • $\sin(B) = 0.9976 \implies \angle B = \sin^{-1}(0.9976) \approx 86^\circ$

  • $\tan(A) = 0.6375 \implies \angle A = \tan^{-1}(0.6375) \approx 33^\circ$

  • $\cos(A) = 0.3432 \implies \angle A = \cos^{-1}(0.3432) \approx 70^\circ$

  • $\tan(D) = 0.3353 \implies \angle D = \tan^{-1}(0.3353) \approx 19^\circ$

• Worked Examples: Solving for Unknown Angles:

  • Problem A: Find $\angle A$ in right triangle $\Delta ABC$ with hypotenuse $AC = 17\,\text{cm}$ and opposite side $BC = 15\,\text{cm}$.     $\sin(A) = \frac{15}{17}$     $\angle A = \sin^{-1}\left(\frac{15}{17}\right) \approx 62^\circ$

  • Problem B: Find $\angle D$ in right triangle $\Delta DEF$ with hypotenuse $DF = 157\,\text{cm}$ and adjacent side $DE = 121\,\text{cm}$.     $\cos(D) = \frac{121}{157}$     $\angle D = \cos^{-1}\left(\frac{121}{157}\right) \approx 40^\circ$

  • Problem C: Find $\angle F$ in right triangle $\Delta DEF$ with opposite side $DE = 37\,\text{cm}$ and adjacent side $EF = 14\,\text{cm}$.     $\tan(F) = \frac{37}{14}$     $\angle F = \tan^{-1}\left(\frac{37}{14}\right) \approx 69^\circ$

  • Problem D: Find $\angle Y$ in right triangle $\Delta WXY$ with hypotenuse $WY = 61\,\text{cm}$ and adjacent side $XY = 45\,\text{cm}$.     $\cos(Y) = \frac{45}{61}$     $\angle Y = \cos^{-1}\left(\frac{45}{61}\right) \approx 42^\circ$

• Composite Triangle Problems:

  • Finding side $AD$ in a connected two-triangle configuration:

  • Given adjacent horizontal side $BC = 13\,\text{cm}$ and angle \angle CAD = 20^\circ$.\n - Using cosine ratio:\n      \cos(20^\circ) = \frac{13}{AD}\n      AD = \frac{13}{\cos(20^\circ)} \approx 13.83\,\text{cm}\n\n# Real-World Applications of Trigonometric Ratios\n\n- **Key Terminology**:\n - **Angle of Elevation**: The angle between the horizontal plane and the line of sight when looking up at an object above eye level.\n - **Angle of Depression**: The angle between the horizontal plane and the line of sight when looking down at an object below eye level.\n\n- **Real-World Word Problems**:\n - **Airplane Altitude Problem**:\n - An air traffic control observation deck is 50\,\text{m}$above ground level. An airplane flies at an altitude of$850\,\text{m} above ground level.\n - Relative height above observation deck:\n      850\,\text{m} - 50\,\text{m} = 800\,\text{m}\n - Angle of elevation from observation deck to plane: 21.5^\circ$.

  • Calculation of direct line-of-sight distance ($d$):       $\sin(21.5^\circ) = \frac{800}{d}$       $d = \frac{800}{\sin(21.5^\circ)} \approx 2182.7\,\text{m}$

  • Silo Height Problem:

  • Distance from observer to silo base: $40\,\text{m}$.

  • Angle of elevation to silo top: 14^\circ$.\n - Calculation of height (h):\n      \tan(14^\circ) = \frac{h}{40}\n      h = 40 \times \tan(14^\circ) \approx 10.0\,\text{m}\n\n - **Archery Distance Problem**:\n - Bullseye is positioned 50\,\text{cm} below archer's eye level.\n - Angle of depression: 5^\circ$.

  • Calculation of flight distance ($d$):       $\sin(5^\circ) = \frac{50}{d}$       $d = \frac{50}{\sin(5^\circ)} \approx 573.7\,\text{cm} \approx 5.74\,\text{m}$

  • Lighthouse and Sailboat Problem:

  • Top of lighthouse: $100\,\text{m}$ above sea level.

  • Angle of depression to sailboat deck: 28^\circ$.\n - Calculation of horizontal distance (x):\n      \tan(28^\circ) = \frac{100}{x}\n      x = \frac{100}{\tan(28^\circ)} \approx 188.1\,\text{m}\n\n# The Sine Law\n\n- **Fundamental Principles**:\n - Applies to acute and oblique (non-right) triangles.\n - Interior angle sum rule for any triangle:\n    A + B + C = 180^\circ\n - **Acute Triangle Definition**: A triangle in which all three interior angles measure less than 90^\circ$.

• The Sine Law Equation:

  • Statement: In any triangle, the ratio of each side length to the sine of its opposite angle is equal.     $\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$

  • Alternative reciprocal form (useful for finding angles):     $\frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}$

• Conditions for Applying the Sine Law:

  1. Given two sides and an angle opposite one of the known sides (SSA).

  2. Given two angles and any side length (AAS or ASA).

• Worked Examples: The Sine Law:

  • Finding Side Lengths:

  • Example A: Triangle $\Delta PQR$ with $\angle P = 68^\circ$, $\angle R = 57^\circ$, and side $q = 8\,\text{cm}$. Find side $r$.

    • Determine third angle $\angle Q$:         $\angle Q = 180^\circ - (68^\circ + 57^\circ) = 55^\circ$

    • Apply Sine Law:         $\frac{r}{\sin(57^\circ)} = \frac{8}{\sin(55^\circ)}$         $r = \frac{8 \times \sin(57^\circ)}{\sin(55^\circ)} \approx 8.2\,\text{cm}$

  • Example B: Triangle $\Delta XYZ$ with $\angle X = 58^\circ$, $\angle Y = 48^\circ$, and side $x = 15\,\text{m}$. Find side $y$.       $\frac{y}{\sin(48^\circ)} = \frac{15}{\sin(58^\circ)}$       $y = \frac{15 \times \sin(48^\circ)}{\sin(58^\circ)} \approx 13.1\,\text{m}$

  • Finding Angle Measures:

  • Example A: Triangle $\Delta ABC$ with $\angle B = 80^\circ$, side $b = 18\,\text{cm}$, side $a = 16\,\text{cm}$. Find $\angle A$.       $\frac{\sin(A)}{16} = \frac{\sin(80^\circ)}{18}$       $\sin(A) = \frac{16 \times \sin(80^\circ)}{18} \approx 0.8754$       $\angle A = \sin^{-1}(0.8754) \approx 61^\circ$

  • Solving Complete Triangles (Finding All Unknown Sides and Angles):

  • Example A: Solve $\Delta STU$ given $\angle S = 54^\circ$, $\angle U = 51^\circ$, side $t = 11\,\text{m}$.

    • Angle $T$:         $\angle T = 180^\circ - (54^\circ + 51^\circ) = 75^\circ$

    • Side $s$:         $\frac{s}{\sin(54^\circ)} = \frac{11}{\sin(75^\circ)} \implies s = \frac{11 \times \sin(54^\circ)}{\sin(75^\circ)} \approx 9.2\,\text{m}$

    • Side $u$:         $\frac{u}{\sin(51^\circ)} = \frac{11}{\sin(75^\circ)} \implies u = \frac{11 \times \sin(51^\circ)}{\sin(75^\circ)} \approx 8.9\,\text{m}$

  • Example B: Solve $\Delta DEF$ given $\angle D = 63^\circ$, side $d = 14\,\text{m}$, side $f = 10\,\text{m}$.

    • Angle $F$:         $\frac{\sin(F)}{10} = \frac{\sin(63^\circ)}{14} \implies \sin(F) = \frac{10 \times \sin(63^\circ)}{14} \approx 0.6364 \implies \angle F \approx 40^\circ$

    • Angle $E$:         $\angle E = 180^\circ - (63^\circ + 40^\circ) = 77^\circ$

    • Side $e$:         $\frac{e}{\sin(77^\circ)} = \frac{14}{\sin(63^\circ)} \implies e = \frac{14 \times \sin(77^\circ)}{\sin(63^\circ)} \approx 15.3\,\text{m}$

  • Triangular Garden Problem:

  • Angela builds a triangular garden with side lengths $15\,\text{m}$ and $13\,\text{m}$. The angle opposite the $15\,\text{m}$ side is 52^\circ$.\n - **Part A**: Angle formed by fence and 15\,\text{m}$side ($\theta):\n      \frac{\sin(\theta)}{13} = \frac{\sin(52^\circ)}{15} \implies \sin(\theta) = \frac{13 \times \sin(52^\circ)}{15} \approx 0.6829 \implies \theta \approx 43^\circ\n - **Part B**: Fence length (f):\n - Third angle = 180^\circ - (52^\circ + 43^\circ) = 85^\circ\n      \frac{f}{\sin(85^\circ)} = \frac{15}{\sin(52^\circ)} \implies f = \frac{15 \times \sin(85^\circ)}{\sin(52^\circ)} \approx 19.0\,\text{m}\n\n# The Cosine Law\n\n- **Definition and Scope**:\n - Relates the three side lengths of a triangle to the cosine of one of its angles.\n - Used for non-right acute or obtuse triangles.\n\n- **Conditions for Applying the Cosine Law**:\n 1. Given **two sides and the contained angle** (SAS - the angle between the two given sides).\n 2. Given **all three side lengths** and seeking an unknown angle (SSS).\n\n\n\n- **The Cosine Law Formulas**:\n - **Formulas for Side Lengths**:\n    a^2 = b^2 + c^2 - 2bc\cos(A)\n    b^2 = a^2 + c^2 - 2ac\cos(B)\n    c^2 = a^2 + b^2 - 2ab\cos(C)\n - **Rearranged Formulas for Angles**:\n    \cos(A) = \frac{b^2 + c^2 - a^2}{2bc}\n    \cos(B) = \frac{a^2 + c^2 - b^2}{2ac}\n    \cos(C) = \frac{a^2 + b^2 - c^2}{2ab}\n\n- **Worked Examples: The Cosine Law**:\n - **Example 1: Finding an Unknown Side**:\n - Determine side length c$in$\Delta ABC$with$a = 7\,\text{cm}$,$b = 5\,\text{cm}$, and$\angle C = 43^\circ$.       $c^2 = 5^2 + 7^2 - 2(5)(7)\cos(43^\circ)$       $c^2 = 25 + 49 - 70(0.7314) = 74 - 51.198 = 22.802$       $c = \sqrt{22.802} \approx 4.8\,\text{cm}$

  • Example 2: Finding an Unknown Angle:

  • Determine $\angle R$ in $\Delta PQR$ with $p = 5.9\,\text{km}$, $q = 6.2\,\text{km}$, and $r = 3.2\,\text{km}$.       $\cos(R) = \frac{5.9^2 + 6.2^2 - 3.2^2}{2(5.9)(6.2)}$       $\cos(R) = \frac{34.81 + 38.44 - 10.24}{73.16} = \frac{63.01}{73.16} \approx 0.8613$       $\angle R = \cos^{-1}(0.8613) \approx 31^\circ$

  • Example 3: Solving a Triangle with SAS:

  • Solve $\Delta ABC$ with $\angle A = 35^\circ$, $b = 9.3\,\text{cm}$, and $c = 12.5\,\text{cm}$.

    • Step 1: Calculate side $a$:         $a^2 = 9.3^2 + 12.5^2 - 2(9.3)(12.5)\cos(35^\circ)$         $a^2 = 86.49 + 156.25 - 232.5(0.8192) = 242.74 - 190.46 = 52.28$         $a = \sqrt{52.28} \approx 7.2\,\text{cm}$

    • Step 2: Calculate $\angle B$ using Sine Law:         $\frac{\sin(B)}{9.3} = \frac{\sin(35^\circ)}{7.2} \implies \sin(B) = \frac{9.3 \times \sin(35^\circ)}{7.2} \approx 0.7408 \implies \angle B \approx 48^\circ$

    • Step 3: Calculate $\angle C$:         $\angle C = 180^\circ - (35^\circ + 48^\circ) = 97^\circ$

  • Example 4: Bruce Trail Hikers Problem:

  • Two hikers set out from a marked tree along paths forming a $50^\circ$ angle. After 2 hours, Hiker 1 is $6\,\text{km}$ away and Hiker 2 is $9\,\text{km}$ away. Find separation distance ($d$).       $d^2 = 6^2 + 9^2 - 2(6)(9)\cos(50^\circ)$       $d^2 = 36 + 81 - 108(0.6428) = 117 - 69.42 = 47.58$       $d = \sqrt{47.58} \approx 6.9\,\text{km}$

Decision Making in Trigonometry

• Method Selection Summary Matrix:

• Method Identification Practice:

  • Case A: Side $c = 30$, $\angle B = 62^\circ$, $\angle C = 35^\circ$. Find side b$.\n - **Selected Method**: Sine Law (Given AAS).\n - Calculation:\n      \frac{b}{\sin(62^\circ)} = \frac{30}{\sin(35^\circ)} \implies b = \frac{30 \times \sin(62^\circ)}{\sin(35^\circ)} \approx 46.2\n - **Case B**: Right triangle with hypotenuse PR = 124$, side$PQ = 74$. Find$\angle R$.

  • Selected Method: Primary Ratio SOH CAH TOA ($\sin$).

  • Calculation:       $\sin(R) = \frac{74}{124} \implies \angle R = \sin^{-1}\left(\frac{74}{124}\right) \approx 37^\circ$

  • Case C: Triangle with sides $DE = 24$, $DF = 20$, $EF = 35$. Find \angle F$.\n - **Selected Method**: Cosine Law (Given SSS).\n - Calculation:\n      \cos(F) = \frac{20^2 + 35^2 - 24^2}{2(20)(35)} = \frac{1049}{1400} \approx 0.7493 \implies \angle F \approx 41^\circ\n - **Case D**: Right triangle with angle V = 29^\circ$, opposite side$TU = 14.3$. Find hypotenuse$u$.

  • Selected Method: Primary Ratio SOH CAH TOA ($\sin$).

  • Calculation:       $\sin(29^\circ) = \frac{14.3}{u} \implies u = \frac{14.3}{\sin(29^\circ)} \approx 29.5$

• Comprehensive Real-World Decision Problems:

  • David's Road Trip (Edmonton to Toronto via Chicago):

  • Route details: Edmonton ($E$) to Chicago ($C$) to Toronto ($T$).

  • Given parameters: Direct distance $ET = 2000\,\text{km}$, $\angle T = 45^\circ$, \angle C = 95^\circ$.\n - Step 1: Calculate angle at Edmonton (\angle E):\n      \angle E = 180^\circ - (45^\circ + 95^\circ) = 40^\circ\n - Step 2: Use Sine Law to find distance Chicago to Toronto (CT = e):\n      \frac{e}{\sin(40^\circ)} = \frac{2000}{\sin(95^\circ)} \implies e = \frac{2000 \times \sin(40^\circ)}{\sin(95^\circ)} \approx 1290.5\,\text{km}\n - Step 3: Use Sine Law to find distance Edmonton to Chicago (EC = t):\n      \frac{t}{\sin(45^\circ)} = \frac{2000}{\sin(95^\circ)} \implies t = \frac{2000 \times \sin(45^\circ)}{\sin(95^\circ)} \approx 1419.6\,\text{km}\n - Step 4: Calculate total driving distance and extra distance:\n      \text{Total Distance} = 1419.6 + 1290.5 = 2710.1\,\text{km}\n      \text{Extra Distance} = 2710.1 - 2000 = 710.1\,\text{km}\n - Therefore, David drove 710.1\,\text{km} further than necessary.\n\n - **Outdoor Hockey Rink Goal Shot**:\n - Goal line width: 5\,\text{ft}.\n - Distance to post 1: 5\,\text{yds} = 15\,\text{ft}.\n - Distance to post 2: 6\,\text{yds} = 18\,\text{ft}.\n - Solve for shooting angle \theta using Cosine Law (SSS):\n      \cos(\theta) = \frac{15^2 + 18^2 - 5^2}{2(15)(18)} = \frac{225 + 324 - 25}{540} = \frac{524}{540} \approx 0.9704\n      \theta = \cos^{-1}(0.9704) \approx 14.0^\circ\n - Therefore, Jill must make her shot within an angle of 14^\circ to hit the net.\n\n - **Evergreen Tree Height Problem**:\n - Initial angle of elevation: 15^\circ$.

  • Walked $31.4\,\text{ft}$ closer to tree.

  • New angle of elevation: 17^\circ$.\n - Step 1: Analyze non-right triangle formed between observation points A$,$B$, and tree top$T:\n - Angle at B$interior to triangle:$180^\circ - 17^\circ = 163^\circ\n - Angle at top T$:$180^\circ - (15^\circ + 163^\circ) = 2^\circ\n - Length AB = 31.4\,\text{ft}\n - Step 2: Use Sine Law to calculate hypotenuse BT = a of the right triangle:\n      \frac{a}{\sin(15^\circ)} = \frac{31.4}{\sin(2^\circ)} \implies a = \frac{31.4 \times \sin(15^\circ)}{\sin(2^\circ)} \approx 232.88\,\text{ft}\n - Step 3: Calculate tree height (h) in the right triangle:\n      \sin(17^\circ) = \frac{h}{232.88} \implies h = 232.88 \times \sin(17^\circ) \approx 68.1\,\text{ft}\n - Therefore, the height of the evergreen tree is approximately 68.1\,\text{ft}$$.