Grade 10 Mathematics Study Guide: Geometric Transformations, Laws of Sines and Cosines, and Bearings

Geometric Transformations: Translation, Reflection, and Rotation

  • Rigid Transformations Overview

    • Rigid transformations alter the position or orientation of a geometric figure on a coordinate plane without changing its shape or size.
    • The three fundamental types of rigid transformations are translation, reflection, and rotation.

  • Translation (Activity No. 3, Part 1 - Mr. Joselito G. Colo)

    • Definition: A translation shifts every point of a geometric figure by the same distance in a specified direction.
    • General Coordinate Mapping Rule:
      • (x,y)→(x+h,y+k)(x, y) \rightarrow (x + h, y + k)
      • Horizontal shift: hh units
      • Vertical shift: kk units
    • Problem Specification:
      • Find the image of quadrilateral RULE\text{RULE} under the translation rule (x,y)→(x−3,y+4)(x, y) \rightarrow (x - 3, y + 4).
    • Original Vertices of Quadrilateral RULE\text{RULE}:
      • R(−1,−1)R(-1, -1)
      • U(3,1)U(3, 1)
      • L(3,−3)L(3, -3)
      • E(−1,−3)E(-1, -3)
    • Step-by-Step Coordinate Transformations for Image R′U′L′E′R'U'L'E':
      • Vertex R′R':
        • x′=−1−3=−4x' = -1 - 3 = -4
        • y′=−1+4=3y' = -1 + 4 = 3
        • R′(−4,3)R'(-4, 3)
      • Vertex U′U':
        • x′=3−3=0x' = 3 - 3 = 0
        • y′=1+4=5y' = 1 + 4 = 5
        • U′(0,5)U'(0, 5)
      • Vertex L′L':
        • x′=3−3=0x' = 3 - 3 = 0
        • y′=−3+4=1y' = -3 + 4 = 1
        • L′(0,1)L'(0, 1)
      • Vertex E′E':
        • x′=−1−3=−4x' = -1 - 3 = -4
        • y′=−3+4=1y' = -3 + 4 = 1
        • E′(−4,1)E'(-4, 1)
    • Graphing Directive:
      • Plot quadrilateral RULE\text{RULE} with vertices R(−1,−1)R(-1, -1), U(3,1)U(3, 1), L(3,−3)L(3, -3), and E(−1,−3)E(-1, -3).
      • Plot the translated image quadrilateral R′U′L′E′R'U'L'E' with vertices R′(−4,3)R'(-4, 3), U′(0,5)U'(0, 5), L′(0,1)L'(0, 1), and E′(−4,1)E'(-4, 1).

  • Reflection (Activity No. 3, Part 2 - Mr. Joselito G. Colo)

    • Definition: A reflection flips a figure across a specified line of reflection, creating a mirror image.
    • Original Figure:
      • Triangle ΔERA\Delta ERA with vertices E(2,6)E(2, 6), R(4,7)R(4, 7), and A(3,3)A(3, 3).
    • Part 2a: Reflection Across the y-axis
      • Coordinate Mapping Rule: (x,y)→(−x,y)(x, y) \rightarrow (-x, y)
      • Reflected Vertices (ΔE′R′A′\Delta E'R'A'):
        • E′(−2,6)E'(-2, 6)
        • R′(−4,7)R'(-4, 7)
        • A′(−3,3)A'(-3, 3)
    • Part 2b: Reflection Across the x-axis
      • Coordinate Mapping Rule: (x,y)→(x,−y)(x, y) \rightarrow (x, -y)
      • Reflected Vertices (ΔE′′R′′A′′\Delta E''R''A''):
        • E′′(2,−6)E''(2, -6)
        • R′′(4,−7)R''(4, -7)
        • A′′(3,−3)A''(3, -3)
    • Graphing Directive:
      • Draw original triangle ΔERA\Delta ERA and its reflection images ΔE′R′A′\Delta E'R'A' across both the y-axis and x-axis on the coordinate plane.

  • Rotation (Activity No. 3, Part 3 - Mr. Joselito G. Colo)

    • Definition: A rotation turns a figure through a specified angle around a fixed center point CC
    • Problem Specification:
      • Draw the image of each given geometric figure for a 120∘120^\circ rotation counterclockwise about center point CC
    • Figures to Rotate:
      • Point PP: Rotate 120∘120^\circ about point CC to obtain point P′P'
      • Line segment ABAB: Rotate 120∘120^\circ about point CC to obtain segment A′B′A'B'
      • Triangle ΔPQR\Delta PQR: Rotate 120∘120^\circ about point CC to obtain triangle ΔP′Q′R′\Delta P'Q'R'

Laws of Sines and Cosines for Oblique Triangles

  • The Law of Sines

    • Formula:
      • asin⁡(A)=bsin⁡(B)=csin⁡(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}
    • Applicability: Used to solve oblique (non-right) triangles when given:
      • Angle-Angle-Side (AAS\text{AAS}
      • Angle-Side-Angle (ASA\text{ASA}
      • Side-Side-Angle (SSA\text{SSA}
    • Triangle Angle Sum Theorem:
      • ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ

  • Fully Worked Example 1: Solving an Oblique Triangle

    • Given:
      • ∠A=59.1∘\angle A = 59.1^\circ
      • ∠C=45.1∘\angle C = 45.1^\circ
      • b=11.5 inb = 11.5\,\text{in}
    • Required:
      • ∠B\angle B
      • Side aa
      • Side cc
    • Solution Procedure:
      • Step 1: Solve for ∠B\angle B
        • ∠B=180∘−(∠A+∠C)\angle B = 180^\circ - (\angle A + \angle C)
        • ∠B=180∘−(59.1∘+45.1∘)\angle B = 180^\circ - (59.1^\circ + 45.1^\circ)
        • ∠B=180∘−104.2∘\angle B = 180^\circ - 104.2^\circ
        • ∠B=75.8∘\angle B = 75.8^\circ
      • Step 2: Solve for side aa using Law of Sines
        • asin⁡(A)=bsin⁡(B)\frac{a}{\sin(A)} = \frac{b}{\sin(B)}
        • asin⁡(59.1∘)=11.5sin⁡(75.8∘)\frac{a}{\sin(59.1^\circ)} = \frac{11.5}{\sin(75.8^\circ)}
        • a=11.5×sin⁡(59.1∘)sin⁡(75.8∘)a = \frac{11.5 \times \sin(59.1^\circ)}{\sin(75.8^\circ)}
        • a=10.18 ina = 10.18\,\text{in}
      • Step 3: Solve for side cc using Law of Sines
        • csin⁡(C)=bsin⁡(B)\frac{c}{\sin(C)} = \frac{b}{\sin(B)}
        • csin⁡(45.1∘)=11.5sin⁡(75.8∘)\frac{c}{\sin(45.1^\circ)} = \frac{11.5}{\sin(75.8^\circ)}
        • c=11.5×sin⁡(45.1∘)sin⁡(75.8∘)c = \frac{11.5 \times \sin(45.1^\circ)}{\sin(75.8^\circ)}
        • c=8.40 inc = 8.40\,\text{in}

  • Fully Worked Example 2: AAS Triangle Case

    • Given:
      • In ΔABC\Delta ABC: ∠A=63∘\angle A = 63^\circ, ∠B=79∘\angle B = 79^\circ, a=12 ina = 12\,\text{in}
    • Classification: This case is an example of AAS\text{AAS} (Angle-Angle-Side).
    • Required:
      • Measure of third angle ∠C\angle C
      • Length of side bb
      • Length of side cc
    • Solution Procedure:
      • Step 1: Solve for ∠C\angle C
        • ∠C=180∘−(∠A+∠B)\angle C = 180^\circ - (\angle A + \angle B)
        • ∠C=180∘−(63∘+79∘)\angle C = 180^\circ - (63^\circ + 79^\circ)
        • ∠C=180∘−142∘\angle C = 180^\circ - 142^\circ
        • ∠C=38∘\angle C = 38^\circ
      • Step 2: Solve for side bb using Law of Sines
        • bsin⁡(B)=asin⁡(A)\frac{b}{\sin(B)} = \frac{a}{\sin(A)}
        • bsin⁡(79∘)=12sin⁡(63∘)\frac{b}{\sin(79^\circ)} = \frac{12}{\sin(63^\circ)}
        • b=12×sin⁡(79∘)sin⁡(63∘)b = \frac{12 \times \sin(79^\circ)}{\sin(63^\circ)}
        • b=13.22 inb = 13.22\,\text{in}

  • The Law of Cosines

    • Formulas for Side Lengths:
      • a2=b2+c2−2×b×c×cos⁡(A)a^2 = b^2 + c^2 - 2 \times b \times c \times \cos(A)
      • b2=a2+c2−2×a×c×cos⁡(B)b^2 = a^2 + c^2 - 2 \times a \times c \times \cos(B)
      • c2=a2+b2−2×a×b×cos⁡(C)c^2 = a^2 + b^2 - 2 \times a \times b \times \cos(C)
    • Formulas for Angles:
      • cos⁡(A)=b2+c2−a22×b×c\cos(A) = \frac{b^2 + c^2 - a^2}{2 \times b \times c}
      • cos⁡(B)=a2+c2−b22×a×c\cos(B) = \frac{a^2 + c^2 - b^2}{2 \times a \times c}
      • cos⁡(C)=a2+b2−c22×a×b\cos(C) = \frac{a^2 + b^2 - c^2}{2 \times a \times b}
    • Applicability: Used when given:
      • Side-Angle-Side (SAS\text{SAS}
      • Side-Side-Side (SSS\text{SSS}

  • Three-Figure True Bearings

    • Measurement Standard: Bearings are measured clockwise from True North (0∘0^\circ or 360∘360^\circ).
    • Cardinal Compass Directions:
      • North (N\text{N}): 000∘000^\circ or 360∘360^\circ
      • East (E\text{E}): 090∘090^\circ
      • South (S\text{S}): 180∘180^\circ
      • West (W\text{W}): 270∘270^\circ

  • Quadrant Bearings

    • Notation Format: Written as N/S θ E/W\text{N/S}\,\theta\,\text{E/W}, where θ\theta is an angle between 0∘0^\circ and 90∘90^\circ
    • Conversion Rules from True Bearings:
      • Range 000∘−090∘000^\circ - 090^\circ (Quadrant 1): N θ E\text{N}\,\theta\,\text{E}
      • Range 090∘−180∘090^\circ - 180^\circ (Quadrant 2): S (180∘−θ) E\text{S}\,(180^\circ - \theta)\,\text{E}
      • Range 180∘−270∘180^\circ - 270^\circ (Quadrant 3): S (θ−180∘) W\text{S}\,(\theta - 180^\circ)\,\text{W}
      • Range 270∘−360∘270^\circ - 360^\circ (Quadrant 4): N (360∘−θ) W\text{N}\,(360^\circ - \theta)\,\text{W}
    • Standard Quadrant Bearing Examples:
      • 1. Point AA from OO: Angle 40∘40^\circ West of North →N 40∘ W\rightarrow \text{N}\,40^\circ\,\text{W}
      • 2. Point BB from OO: Angle 35∘35^\circ East of North →N 35∘ E\rightarrow \text{N}\,35^\circ\,\text{E}
      • 3. Point CC from OO: Angle 73∘73^\circ West of South →S 73∘ W\rightarrow \text{S}\,73^\circ\,\text{W}
      • 4. Point DD from OO: True bearing 126∘126^\circ. Angle from South: 180∘−126∘=54∘→S 54∘ E180^\circ - 126^\circ = 54^\circ \rightarrow \text{S}\,54^\circ\,\text{E}

Grade 10 Mathematics Activity Sheet - Complete Solutions

  • Part I: Law of Sines Problems

    • Problem 1:
      • Given: In ΔABC\Delta ABC, ∠A=42∘\angle A = 42^\circ, ∠B=68∘\angle B = 68^\circ, side a=12 cma = 12\,\text{cm}.
      • Required: Find side bb
      • Calculation:bsin⁡(68∘)=12sin⁡(42∘)\frac{b}{\sin(68^\circ)} = \frac{12}{\sin(42^\circ)}b = \frac{12 \times \sin(68^\circ)}{\sin(42^\circ)}b=12×0.92720.6691b = \frac{12 \times 0.9272}{0.6691}b \approx 16.63\,\text{cm}\n * **Problem 2:**\n * *Given:* A triangle has \angle A = 35^\circ,,a = 18\,\text{m},,b = 25\,\text{m}.\n * *Required:* Find \angle B\n * *Calculation:*\frac{\sin(B)}{25} = \frac{\sin(35^\circ)}{18}sin⁡(B)=25×sin⁡(35∘)18\sin(B) = \frac{25 \times \sin(35^\circ)}{18}\sin(B) = \frac{25 \times 0.5736}{18}sin⁡(B)≈0.7966\sin(B) \approx 0.7966\angle B = \arcsin(0.7966) \approx 52.80^\circ\n * **Problem 3:**\n * *Given:* A surveyor measures two angles of a triangular field: \angle A = 50^\circ,,\angle B = 60^\circ,side, sidea = 45\,\text{m}.\n * *Required:* Find side b\n * *Calculation:*\frac{b}{\sin(60^\circ)} = \frac{45}{\sin(50^\circ)}b=45×sin⁡(60∘)sin⁡(50∘)b = \frac{45 \times \sin(60^\circ)}{\sin(50^\circ)}b = \frac{45 \times 0.8660}{0.7660}b≈50.87 mb \approx 50.87\,\text{m}
    • Problem 4:
      • Given: A tree casts a shadow forming a triangle with ∠A=40∘\angle A = 40^\circ, ∠B=85∘\angle B = 85^\circ, side b=30 mb = 30\,\text{m}.
      • Required: Find side aa
      • Calculation:asin⁡(40∘)=30sin⁡(85∘)\frac{a}{\sin(40^\circ)} = \frac{30}{\sin(85^\circ)}a = \frac{30 \times \sin(40^\circ)}{\sin(85^\circ)}a=30×0.64280.9962a = \frac{30 \times 0.6428}{0.9962}a \approx 19.36\,\text{m}\n * **Problem 5:**\n * *Given:* A rescue tower observes two boats with a = 18\,\text{km},,\angle A = 45^\circ,,\angle B = 70^\circ.\n * *Required:* Find side b\n * *Calculation:*\frac{b}{\sin(70^\circ)} = \frac{18}{\sin(45^\circ)}b=18×sin⁡(70∘)sin⁡(45∘)b = \frac{18 \times \sin(70^\circ)}{\sin(45^\circ)}b = \frac{18 \times 0.9397}{0.7071}b≈23.92 kmb \approx 23.92\,\text{km}

  • Part II: Law of Cosines Problems

    • Problem 6:
      • Given: a=8 cma = 8\,\text{cm}, b=12 cmb = 12\,\text{cm}, ∠C=60∘\angle C = 60^\circ
      • Required: Find side cc
      • Calculation:c2=a2+b2−2×a×b×cos⁡(C)c^2 = a^2 + b^2 - 2 \times a \times b \times \cos(C)c^2 = 8^2 + 12^2 - 2 \times 8 \times 12 \times \cos(60^\circ)c2=64+144−192×0.5c^2 = 64 + 144 - 192 \times 0.5c^2 = 208 - 96 = 112c=112≈10.58 cmc = \sqrt{112} \approx 10.58\,\text{cm}
    • Problem 7:
      • Given: b=14 mb = 14\,\text{m}, c=18 mc = 18\,\text{m}, ∠A=70∘\angle A = 70^\circ
      • Required: Find side aa
      • Calculation:a2=b2+c2−2×b×c×cos⁡(A)a^2 = b^2 + c^2 - 2 \times b \times c \times \cos(A)a^2 = 14^2 + 18^2 - 2 \times 14 \times 18 \times \cos(70^\circ)a2=196+324−504×0.3420a^2 = 196 + 324 - 504 \times 0.3420a^2 = 520 - 172.38 = 347.62a=347.62≈18.64 ma = \sqrt{347.62} \approx 18.64\,\text{m}
    • Problem 8:
      • Given: A triangular park has sides 12 m12\,\text{m}, 15 m15\,\text{m}, and 18 m18\,\text{m}.
      • Required: Find the largest angle.
      • Calculation: The largest angle lies opposite the longest side (c=18 mc = 18\,\text{m}).cos⁡(C)=a2+b2−c22×a×b\cos(C) = \frac{a^2 + b^2 - c^2}{2 \times a \times b}\cos(C) = \frac{12^2 + 15^2 - 18^2}{2 \times 12 \times 15}cos⁡(C)=144+225−324360\cos(C) = \frac{144 + 225 - 324}{360}\cos(C) = \frac{45}{360} = 0.125∠C=arccos⁡(0.125)≈82.82∘\angle C = \arccos(0.125) \approx 82.82^\circ
    • Problem 9:
      • Given: Two roads meet at an angle of 95∘95^\circ. One road is 8 km8\,\text{km} long, the other is 11 km11\,\text{km} long.
      • Required: Find the distance between their endpoints.
      • Calculation:d2=82+112−2×8×11×cos⁡(95∘)d^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos(95^\circ)d^2 = 64 + 121 - 176 \times (-0.08716)d2=185+15.34=200.34d^2 = 185 + 15.34 = 200.34d = \sqrt{200.34} \approx 14.15\,\text{km}\n * **Problem 10:**\n * *Given:* A surveyor measures two sides of a lot: 35\,\text{m}andand42\,\text{m},withanincludedangleof, with an included angle of58^\circ\n * *Required:* Find the third side.\n * *Calculation:*x^2 = 35^2 + 42^2 - 2 \times 35 \times 42 \times \cos(58^\circ)x2=1225+1764−2940×0.5299x^2 = 1225 + 1764 - 2940 \times 0.5299x^2 = 2989 - 1557.96 = 1431.04x=1431.04≈37.83 mx = \sqrt{1431.04} \approx 37.83\,\text{m}

  • Part III: Bearings and Navigation Problems

    • Problem 11:
      • Given: A boat travels 15 km15\,\text{km} on a bearing of 060∘060^\circ
      • Required: Draw the bearing and determine its direction as a quadrant bearing.
      • Calculation / Expressed Value: True Bearing: 060∘060^\circ
      • Quadrant Bearing: N 60∘ E\text{N}\,60^\circ\,\text{E}
      • Distance: 15 km15\,\text{km}
    • Problem 12:
      • Given: An airplane flies 40 km40\,\text{km} on a bearing of 140∘140^\circ
      • Required: Express the bearing as a quadrant bearing.
      • Calculation:θ=180∘−140∘=40∘\theta = 180^\circ - 140^\circ = 40^\circ
      • Quadrant Bearing: S 40∘ E\text{S}\,40^\circ\,\text{E}
      • Distance: 40 km40\,\text{km}
    • Problem 13:
      • Given: A ship sails 20 km20\,\text{km} due North, then 15 km15\,\text{km} on a bearing of 070∘070^\circ
      • Required: Find its approximate distance from the starting point.
      • Calculation: The interior angle of the triangle formed at the point of turn is 180∘−70∘=110∘180^\circ - 70^\circ = 110^\circ.d2=202+152−2×20×15×cos⁡(110∘)d^2 = 20^2 + 15^2 - 2 \times 20 \times 15 \times \cos(110^\circ)d^2 = 400 + 225 - 600 \times (-0.3420)d2=625+205.21=830.21d^2 = 625 + 205.21 = 830.21d = \sqrt{830.21} \approx 28.81\,\text{km}\n * **Problem 14:**\n * *Given:* A hiker walks 10\,\text{km}East,thenEast, then8\,\text{km}onabearingofon a bearing of330^\circ\n * *Required:* Find the distance from the starting point.\n * *Calculation:* East direction corresponds to bearing 090^\circ.Theinterioranglebetweenthetwovectorsinthetriangleis. The interior angle between the two vectors in the triangle is60^\circ..d^2 = 10^2 + 8^2 - 2 \times 10 \times 8 \times \cos(60^\circ)d2=100+64−160×0.5d^2 = 100 + 64 - 160 \times 0.5d^2 = 164 - 80 = 84d=84≈9.17 kmd = \sqrt{84} \approx 9.17\,\text{km}
    • Problem 15:
      • Given: A lighthouse is located 18 km18\,\text{km} from a ship on a bearing of 120∘120^\circ. Another ship is 25 km25\,\text{km} away on a bearing of 165∘165^\circ
      • Required: Find the distance between the two ships.
      • Calculation: The angle between the two bearing lines from the lighthouse is 165∘−120∘=45∘165^\circ - 120^\circ = 45^\circ.d2=182+252−2×18×25×cos⁡(45∘)d^2 = 18^2 + 25^2 - 2 \times 18 \times 25 \times \cos(45^\circ)d^2 = 324 + 625 - 900 \times 0.7071d2=949−636.40=312.60d^2 = 949 - 636.40 = 312.60d = \sqrt{312.60} \approx 17.68\,\text{km}$$