{"id":980,"date":"2026-01-04T14:44:43","date_gmt":"2026-01-04T04:44:43","guid":{"rendered":"https:\/\/archive4ones.com\/2ndstudy\/?p=980"},"modified":"2026-01-04T18:01:33","modified_gmt":"2026-01-04T08:01:33","slug":"year11-math-2-2-4-warm-up-questions-applied-trigonometry-general-mathematics","status":"publish","type":"post","link":"https:\/\/archive4ones.com\/2ndstudy\/?p=980","title":{"rendered":"Year11 MATH 2-2-4 Warm-up Questions-Applied Trigonometry (General Mathematics)"},"content":{"rendered":"\n

To master Unit 2 of General Mathematics, you must move beyond the right-angled trigonometry of earlier years. Chapter 4 introduces the Sine Rule<\/strong>, the Cosine Rule<\/strong>, and the Area Rule<\/strong>, which allow you to solve for any triangle, anywhere. This skill is critical for advanced navigation, surveying, and 3D architectural modeling.<\/p>\n\n\n\n

Concepts and Skills Covered:<\/strong><\/h3>\n\n\n\n
    \n
  1. The Sine Rule:<\/strong> Using a<\/mi>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi><\/mrow><\/mfrac>=<\/mo>b<\/mi>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>B<\/mi><\/mrow><\/mfrac><\/mrow>\\frac{a}{\\sin A} = \\frac{b}{\\sin B}<\/annotation><\/semantics><\/math> to find missing sides and angles in non-right-angled triangles.<\/li>\n\n\n\n
  2. The Cosine Rule:<\/strong> Applying a<\/mi>2<\/mn><\/msup>=<\/mo>b<\/mi>2<\/mn><\/msup>+<\/mo>c<\/mi>2<\/mn><\/msup>\u2212<\/mo>2<\/mn>b<\/mi>c<\/mi><\/mspace>cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi><\/mrow>a^2 = b^2 + c^2 – 2bc \\cos A<\/annotation><\/semantics><\/math> when provided with two sides and an included angle (SAS) or three sides (SSS).<\/li>\n\n\n\n
  3. Area of a Triangle:<\/strong> Calculating area using Area<\/mtext>=<\/mo>1<\/mn>2<\/mn><\/mfrac>a<\/mi>b<\/mi><\/mspace>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>C<\/mi><\/mrow>\\text{Area} = \\frac{1}{2}ab \\sin C<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
  4. Strategic Selection:<\/strong> Determining which trigonometric rule is most efficient based on the given information.<\/li>\n\n\n\n
  5. Navigation and Bearings:<\/strong> Integrating bearings into trigonometric diagrams.<\/li>\n<\/ol>\n\n\n\n

    Q1. In \u0394<\/mi><\/mrow>A<\/mi>B<\/mi>C<\/mi><\/mrow>\\Delta ABC<\/annotation><\/semantics><\/math>, you are given \u2220<\/mi>A<\/mi>=<\/mo>40<\/mn>\u2218<\/mo><\/msup><\/mrow>\\angle A = 40^{\\circ}<\/annotation><\/semantics><\/math>, \u2220<\/mi>B<\/mi>=<\/mo>60<\/mn>\u2218<\/mo><\/msup><\/mrow>\\angle B = 60^{\\circ}<\/annotation><\/semantics><\/math>, and side b<\/mi>=<\/mo>10<\/mn> cm<\/mtext><\/mrow>b = 10\\text{ cm}<\/annotation><\/semantics><\/math>. Which calculation correctly finds the length of side a<\/mi>a<\/annotation><\/semantics><\/math>?<\/p>\n\n\n\n

    A. a<\/mi>=<\/mo>10<\/mn>\u00d7<\/mo>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>40<\/mn>\u2218<\/mo><\/msup>\u00d7<\/mo>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>60<\/mn>\u2218<\/mo><\/msup><\/mrow>a = 10 \\times \\sin 40^{\\circ} \\times \\sin 60^{\\circ}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    B. a<\/mi>=<\/mo>10<\/mn><\/mspace>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>40<\/mn>\u2218<\/mo><\/msup><\/mrow>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>60<\/mn>\u2218<\/mo><\/msup><\/mrow><\/mfrac><\/mrow>a = \\frac{10 \\sin 40^{\\circ}}{\\sin 60^{\\circ}}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    C. a<\/mi>=<\/mo>10<\/mn><\/mspace>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>60<\/mn>\u2218<\/mo><\/msup><\/mrow>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>40<\/mn>\u2218<\/mo><\/msup><\/mrow><\/mfrac><\/mrow>a = \\frac{10 \\sin 60^{\\circ}}{\\sin 40^{\\circ}}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    D. a<\/mi>2<\/mn><\/msup>=<\/mo>10<\/mn>2<\/mn><\/msup>+<\/mo>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>40<\/mn>\u2218<\/mo><\/msup>\u2212<\/mo>2<\/mn>(<\/mo>10<\/mn>)<\/mo><\/mspace>cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>60<\/mn>\u2218<\/mo><\/msup><\/mrow>a^2 = 10^2 + \\sin 40^{\\circ} – 2(10)\\cos 60^{\\circ}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hint: Recall that the Sine Rule relates side lengths to the sine of the angle directly across from them.<\/p>\n\n\n\n

    Q2. To find a missing angle in a triangle where all three side lengths are known, which trigonometric rule is the most direct to use?<\/p>\n\n\n\n

    A. Sine Rule<\/p>\n\n\n\n

    B. Pythagoras’ Theorem<\/p>\n\n\n\n

    C. SOH CAH TOA<\/p>\n\n\n\n

    D. Cosine Rule<\/p>\n\n\n\n

    Hint: Consider which formula allows you to input three sides to output one angle.<\/p>\n\n\n\n

    Q3. In \u0394<\/mi><\/mrow>P<\/mi>Q<\/mi>R<\/mi><\/mrow>\\Delta PQR<\/annotation><\/semantics><\/math>, p<\/mi>=<\/mo>7<\/mn> cm<\/mtext><\/mrow>p = 7\\text{ cm}<\/annotation><\/semantics><\/math>, q<\/mi>=<\/mo>9<\/mn> cm<\/mtext><\/mrow>q = 9\\text{ cm}<\/annotation><\/semantics><\/math>, and \u2220<\/mi>R<\/mi>=<\/mo>48<\/mn>\u2218<\/mo><\/msup><\/mrow>\\angle R = 48^{\\circ}<\/annotation><\/semantics><\/math>. Which formula would find the length of side r<\/mi>r<\/annotation><\/semantics><\/math> ?<\/p>\n\n\n\n

    A. r<\/mi>=<\/mo>7<\/mn>2<\/mn><\/msup>+<\/mo>9<\/mn>2<\/mn><\/msup><\/mrow><\/msqrt><\/mrow>r = \\sqrt{7^2 + 9^2}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    B. r<\/mi>2<\/mn><\/msup>=<\/mo>7<\/mn>2<\/mn><\/msup>+<\/mo>9<\/mn>2<\/mn><\/msup>\u2212<\/mo>2<\/mn>(<\/mo>7<\/mn>)<\/mo>(<\/mo>9<\/mn>)<\/mo><\/mspace>cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>48<\/mn>\u2218<\/mo><\/msup><\/mrow>r^2 = 7^2 + 9^2 – 2(7)(9)\\cos 48^{\\circ}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    C. r<\/mi>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>48<\/mn>\u2218<\/mo><\/msup><\/mrow><\/mfrac>=<\/mo>7<\/mn>9<\/mn><\/mfrac><\/mrow>\\frac{r}{\\sin 48^{\\circ}} = \\frac{7}{9}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    D. r<\/mi>=<\/mo>7<\/mn><\/mspace>cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>48<\/mn>\u2218<\/mo><\/msup>+<\/mo>9<\/mn><\/mspace>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>48<\/mn>\u2218<\/mo><\/msup><\/mrow>r = 7 \\cos 48^{\\circ} + 9 \\sin 48^{\\circ}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hint: When you have two sides and the angle trapped between them, the Cosine Rule is the appropriate tool.<\/p>\n\n\n\n

    Q4. Calculate the area of \u0394<\/mi><\/mrow>X<\/mi>Y<\/mi>Z<\/mi><\/mrow>\\Delta XYZ<\/annotation><\/semantics><\/math> if x<\/mi>=<\/mo>10<\/mn> cm<\/mtext><\/mrow>x = 10\\text{ cm}<\/annotation><\/semantics><\/math>, y<\/mi>=<\/mo>15<\/mn> cm<\/mtext><\/mrow>y = 15\\text{ cm}<\/annotation><\/semantics><\/math>, and \u2220<\/mi>Z<\/mi>=<\/mo>30<\/mn>\u2218<\/mo><\/msup><\/mrow>\\angle Z = 30^{\\circ}<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    A. 64.9<\/mn> cm<\/mtext>2<\/mn><\/msup><\/mrow>64.9\\text{ cm}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    B. 75<\/mn> cm<\/mtext>2<\/mn><\/msup><\/mrow>75\\text{ cm}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    C. 37.5<\/mn> cm<\/mtext>2<\/mn><\/msup><\/mrow>37.5\\text{ cm}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    D. 150<\/mn> cm<\/mtext>2<\/mn><\/msup><\/mrow>150\\text{ cm}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hint: The formula for the area of a non-right-angled triangle involves sine and the included angle.<\/p>\n\n\n\n

    Q5. When using the Sine Rule to find an angle, which condition might lead to the ‘ambiguous case’ (two possible triangles)?<\/p>\n\n\n\n

    A. When given two angles and the included side (ASA).<\/p>\n\n\n\n

    B. When given two sides and a non-included acute angle (SSA).<\/p>\n\n\n\n

    C. When the triangle is right-angled.<\/p>\n\n\n\n

    D. When given all three side lengths (SSS).<\/p>\n\n\n\n

    Ambiguity arises when a specific set of side and angle information could theoretically draw two different shapes.<\/p>\n\n\n\n

    Q6. A surveyor needs to find the distance between two points, A and B, separated by a lake. They stand at point C and measure A<\/mi>C<\/mi>=<\/mo>120<\/mn> m<\/mtext><\/mrow>AC = 120\\text{ m}<\/annotation><\/semantics><\/math>, B<\/mi>C<\/mi>=<\/mo>150<\/mn> m<\/mtext><\/mrow>BC = 150\\text{ m}<\/annotation><\/semantics><\/math>, and \u2220<\/mi>A<\/mi>C<\/mi>B<\/mi>=<\/mo>40<\/mn>\u2218<\/mo><\/msup><\/mrow>\\angle ACB = 40^{\\circ}<\/annotation><\/semantics><\/math>. What is the first step?<\/p>\n\n\n\n

    A. Calculate the average of the two known sides.<\/p>\n\n\n\n

    B. Use the Sine Rule to find \u2220ABC<\/em>.<\/p>\n\n\n\n

    C. Use the Cosine Rule to find the distance AB<\/em>.<\/p>\n\n\n\n

    D. Assume the triangle is right-angled at C.<\/p>\n\n\n\n

    Hint: Evaluate the information provided: two sides and the angle between them.<\/p>\n\n\n\n

    Q7. If cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi><\/mrow>\\cos A<\/annotation><\/semantics><\/math> is calculated to be a negative value (e.g., \u2212<\/mo>0.25<\/mn><\/mrow>-0.25<\/annotation><\/semantics><\/math>) when using the Cosine Rule, what does this tell you about \u2220<\/mi>A<\/mi><\/mrow>\\angle A<\/annotation><\/semantics><\/math>?<\/p>\n\n\n\n

    <\/p>\n\n\n\n

    A. Angle A is acute (0<\/mn>\u2218<\/mo><\/msup><<\/mo>A<\/mi><<\/mo>90<\/mn>\u2218<\/mo><\/msup><\/mrow>0^{\\circ} < A < 90^{\\circ}<\/annotation><\/semantics><\/math>)..<\/p>\n\n\n\n

    B. Angle A is obtuse (90<\/mn>\u2218<\/mo><\/msup><<\/mo>A<\/mi><<\/mo>180<\/mn>\u2218<\/mo><\/msup><\/mrow>90^{\\circ} < A < 180^{\\circ}<\/annotation><\/semantics><\/math>)..<\/p>\n\n\n\n

    C. The triangle cannot exist.<\/p>\n\n\n\n

    D. The calculation is incorrect.<\/p>\n\n\n\n

    Hint: Recall the behaviour of the cosine graph or the Unit Circle for angles greater than 90<\/mn>90<\/annotation><\/semantics><\/math> degrees.<\/p>\n\n\n\n

    Q8. A ship travels on a bearing of 060<\/mn>\u2218<\/mo><\/msup>060^{\\circ}<\/annotation><\/semantics><\/math> for 20<\/mn> km<\/mtext><\/mrow>20\\text{ km}<\/annotation><\/semantics><\/math>, then turns and travels on a bearing of 150<\/mn>\u2218<\/mo><\/msup>150^{\\circ}<\/annotation><\/semantics><\/math> for 15<\/mn> km<\/mtext><\/mrow>15\\text{ km}<\/annotation><\/semantics><\/math>. To find the direct distance back to the start, what is the internal angle at the turn?<\/p>\n\n\n\n

    A. \u2220=60\u2218<\/p>\n\n\n\n

    B. \u2220=150\u2218<\/p>\n\n\n\n

    C. \u2220=210\u2218<\/p>\n\n\n\n

    D. \u2220=90\u2218<\/p>\n\n\n\n

    Hint: Draw a diagram with ‘North’ lines at each turn to visualize the relationship between the bearings.<\/p>\n\n\n\n

    Q9. Which of these is the correct rearrangement of the Cosine Rule (a<\/mi>2<\/mn><\/msup>=<\/mo>b<\/mi>2<\/mn><\/msup>+<\/mo>c<\/mi>2<\/mn><\/msup>\u2212<\/mo>2<\/mn>b<\/mi>c<\/mi><\/mspace>cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi><\/mrow>a^2 = b^2 + c^2 – 2bc \\cos A<\/annotation><\/semantics><\/math>) to solve for the angle A<\/mi>A<\/annotation><\/semantics><\/math>?<\/p>\n\n\n\n

    A. cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi>=<\/mo>2<\/mn>b<\/mi>c<\/mi><\/mrow>b<\/mi>2<\/mn><\/msup>+<\/mo>c<\/mi>2<\/mn><\/msup>\u2212<\/mo>a<\/mi>2<\/mn><\/msup><\/mrow><\/mfrac><\/mrow>\\cos A = \\frac{2bc}{b^2 + c^2 – a^2}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    B. cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi>=<\/mo>a<\/mi>2<\/mn><\/msup>+<\/mo>b<\/mi>2<\/mn><\/msup>\u2212<\/mo>c<\/mi>2<\/mn><\/msup><\/mrow>2<\/mn>a<\/mi>b<\/mi><\/mrow><\/mfrac><\/mrow>\\cos A = \\frac{a^2 + b^2 – c^2}{2ab}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    C. cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi>=<\/mo>b<\/mi>2<\/mn><\/msup>+<\/mo>c<\/mi>2<\/mn><\/msup>\u2212<\/mo>a<\/mi>2<\/mn><\/msup><\/mrow>2<\/mn>b<\/mi>c<\/mi><\/mrow><\/mfrac><\/mrow>\\cos A = \\frac{b^2 + c^2 – a^2}{2bc}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    D. cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi>=<\/mo>a<\/mi>2<\/mn><\/msup>\u2212<\/mo>b<\/mi>2<\/mn><\/msup>\u2212<\/mo>c<\/mi>2<\/mn><\/msup>\u2212<\/mo>2<\/mn>b<\/mi>c<\/mi><\/mrow>\\cos A = a^2 – b^2 – c^2 – 2bc<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hint: Isolate the term containing ‘cos A’ on one side of the equation first.<\/p>\n\n\n\n

    Q10. A 10<\/mn> m<\/mtext><\/mrow>10\\text{ m}<\/annotation><\/semantics><\/math> ladder leans against a wall. If the ladder makes an angle of 70<\/mn>\u2218<\/mo><\/msup>70^{\\circ}<\/annotation><\/semantics><\/math> with the ground, how high up the wall does it reach?<\/p>\n\n\n\n

    A. 3.42 m<\/p>\n\n\n\n

    B. 9.40 m<\/p>\n\n\n\n

    C. 27.47 m<\/p>\n\n\n\n

    D. 10.64 m<\/p>\n\n\n\n

    Hint: Identify the ladder as the hypotenuse and the height as the opposite side in a right-angled triangle.<\/p>\n\n\n\n

    ****************<\/p>\n\n\n\n

    \u4e00\u822c\u6570\u5b66\u30e6\u30cb\u30c3\u30c82\u3092\u30de\u30b9\u30bf\u30fc\u3059\u308b\u306b\u306f\u3001\u3053\u308c\u307e\u3067\u306e\u76f4\u89d2\u4e09\u89d2\u6cd5\u306e\u67a0\u3092\u8d85\u3048\u306a\u3051\u308c\u3070\u306a\u308a\u307e\u305b\u3093\u3002\u7b2c4\u7ae0\u3067\u306f\u3001\u6b63\u5f26\u5b9a\u7406\u3001\u4f59\u5f26\u5b9a\u7406\u3001\u9762\u7a4d\u5b9a\u7406\u3092\u7d39\u4ecb\u3057\u307e\u3059\u3002\u3053\u308c\u3089\u306e\u5b9a\u7406\u306f\u3001\u3042\u3089\u3086\u308b\u4e09\u89d2\u5f62\u3092\u3001\u3042\u3089\u3086\u308b\u5834\u6240\u3067\u89e3\u304f\u306e\u306b\u5f79\u7acb\u3061\u307e\u3059\u3002\u3053\u306e\u30b9\u30ad\u30eb\u306f\u3001\u9ad8\u5ea6\u306a\u30ca\u30d3\u30b2\u30fc\u30b7\u30e7\u30f3\u3001\u6e2c\u91cf\u30013D\u5efa\u7bc9\u30e2\u30c7\u30ea\u30f3\u30b0\u306b\u4e0d\u53ef\u6b20\u3067\u3059\u3002<\/p>\n\n\n\n

    \u5b66\u7fd2\u5185\u5bb9\u3068\u30b9\u30ad\u30eb\uff1a<\/p>\n\n\n\n

    \u6b63\u5f26\u5b9a\u7406\uff1aa<\/mi>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi><\/mrow><\/mfrac>=<\/mo>b<\/mi>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>B<\/mi><\/mrow><\/mfrac><\/mrow>\\frac{a}{\\sin A} = \\frac{b}{\\sin B}<\/annotation><\/semantics><\/math> \u3092\u7528\u3044\u3066\u3001\u76f4\u89d2\u3067\u306a\u3044\u4e09\u89d2\u5f62\u306e\u6b20\u3051\u3066\u3044\u308b\u8fba\u3068\u89d2\u5ea6\u3092\u6c42\u3081\u307e\u3059\u3002<\/p>\n\n\n\n

    \u4f59\u5f26\u5b9a\u7406\uff1a2\u8fba\u30681\u3064\u306e\u593e\u89d2\uff08SAS\uff09\u307e\u305f\u306f3\u8fba\uff08SSS\uff09\u304c\u4e0e\u3048\u3089\u308c\u3066\u3044\u308b\u5834\u5408\u3001a<\/mi>2<\/mn><\/msup>=<\/mo>b<\/mi>2<\/mn><\/msup>+<\/mo>c<\/mi>2<\/mn><\/msup>\u2212<\/mo>2<\/mn>b<\/mi>c<\/mi><\/mspace>cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi><\/mrow>a^2 = b^2 + c^2 – 2bc \\cos A<\/annotation><\/semantics><\/math> \u3092\u9069\u7528\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

    \u4e09\u89d2\u5f62\u306e\u9762\u7a4d\uff1a\u9762\u7a4d<\/mtext>=<\/mo>1<\/mn>2<\/mn><\/mfrac>a<\/mi>b<\/mi><\/mspace>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>C<\/mi><\/mrow>\\text{\u9762\u7a4d} = \\frac{1}{2}ab \\sin C<\/annotation><\/semantics><\/math> \u3092\u7528\u3044\u3066\u9762\u7a4d\u3092\u8a08\u7b97\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

    \u6226\u7565\u7684\u9078\u629e\uff1a\u4e0e\u3048\u3089\u308c\u305f\u60c5\u5831\u306b\u57fa\u3065\u3044\u3066\u3001\u3069\u306e\u4e09\u89d2\u95a2\u6570\u306e\u898f\u5247\u304c\u6700\u3082\u52b9\u7387\u7684\u304b\u3092\u6c7a\u5b9a\u3059\u308b\u3002<\/p>\n\n\n\n

    \u30ca\u30d3\u30b2\u30fc\u30b7\u30e7\u30f3\u3068\u65b9\u4f4d\uff1a\u65b9\u4f4d\u3092\u4e09\u89d2\u95a2\u6570\u56f3\u306b\u7d44\u307f\u8fbc\u3080\u3002<\/p>\n\n\n\n

    **********************<\/p>\n\n\n\n

    Right answers<\/p>\n\n\n\n

    Q1. B. a<\/mi>=<\/mo>10<\/mn><\/mspace>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>40<\/mn>\u2218<\/mo><\/msup><\/mrow>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>60<\/mn>\u2218<\/mo><\/msup><\/mrow><\/mfrac><\/mrow>a = \\frac{10 \\sin 40^{\\circ}}{\\sin 60^{\\circ}}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Applying the Sine Rule, we set up the ratio a<\/mi>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi><\/mrow><\/mfrac>=<\/mo>b<\/mi>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>B<\/mi><\/mrow><\/mfrac><\/mrow>\\frac{a}{\\sin A} = \\frac{b}{\\sin B}<\/annotation><\/semantics><\/math> and rearrange to solve for a<\/mi>a<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q2. D. Cosine Rule<\/p>\n\n\n\n

    The Cosine Rule can be rearranged to isolate the cosine of an angle, making it the standard choice for ‘Side-Side-Side’ (SSS) scenarios.<\/p>\n\n\n\n

    Q3. B. r<\/mi>2<\/mn><\/msup>=<\/mo>7<\/mn>2<\/mn><\/msup>+<\/mo>9<\/mn>2<\/mn><\/msup>\u2212<\/mo>2<\/mn>(<\/mo>7<\/mn>)<\/mo>(<\/mo>9<\/mn>)<\/mo><\/mspace>cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>48<\/mn>\u2218<\/mo><\/msup><\/mrow>r^2 = 7^2 + 9^2 – 2(7)(9)\\cos 48^{\\circ}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    This is the correct application of the Cosine Rule for the Side-Angle-Side (SAS) configuration.<\/p>\n\n\n\n

    Q4. C. 37.5<\/mn> cm<\/mtext>2<\/mn><\/msup><\/mrow>37.5\\text{ cm}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Using Area<\/mtext>=<\/mo>1<\/mn>2<\/mn><\/mfrac>x<\/mi>y<\/mi><\/mspace>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>Z<\/mi><\/mrow>\\text{Area} = \\frac{1}{2}xy \\sin Z<\/annotation><\/semantics><\/math>, we calculate 0.5<\/mn>\u00d7<\/mo>10<\/mn>\u00d7<\/mo>15<\/mn>\u00d7<\/mo>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>30<\/mn>\u2218<\/mo><\/msup>=<\/mo>37.5<\/mn><\/mrow>0.5 \\times 10 \\times 15 \\times \\sin 30^{\\circ} = 37.5<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q5. B. When given two sides and a non-included acute angle (SSA).<\/p>\n\n\n\n

    The ambiguous case occurs because the sine of an angle \u03b8<\/mi>\\theta<\/annotation><\/semantics><\/math> is the same as the sine of its supplement (180<\/mn>\u2218<\/mo><\/msup>\u2212<\/mo>\u03b8<\/mi><\/mrow>180^{\\circ} – \\theta<\/annotation><\/semantics><\/math>).<\/p>\n\n\n\n

    Q6. C. Use the Cosine Rule to find the distance A<\/mi>B<\/mi><\/mrow>AB<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Since the surveyor has two sides and the included angle (SAS), the Cosine Rule is the only way to find the third side directly.<\/p>\n\n\n\n

    Q7. B. Angle A is obtuse (90<\/mn>\u2218<\/mo><\/msup><<\/mo>A<\/mi><<\/mo>180<\/mn>\u2218<\/mo><\/msup><\/mrow>90^{\\circ} < A < 180^{\\circ}<\/annotation><\/semantics><\/math>).<\/p>\n\n\n\n

    The cosine function is negative in the second quadrant, which corresponds to obtuse angles in a triangle.<\/p>\n\n\n\n

    Q8. D. \u2220<\/mi>=<\/mo>90<\/mn>\u2218<\/mo><\/msup><\/mrow>\\angle = 90^{\\circ}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    The difference between the bearings 150<\/mn>\u2218<\/mo><\/msup>150^{\\circ}<\/annotation><\/semantics><\/math> and the back-bearing of the first leg (60<\/mn>\u2218<\/mo><\/msup>+<\/mo>180<\/mn>\u2218<\/mo><\/msup>=<\/mo>240<\/mn>\u2218<\/mo><\/msup><\/mrow>60^{\\circ} + 180^{\\circ} = 240^{\\circ}<\/annotation><\/semantics><\/math>) or simple geometry shows the internal angle is 90<\/mn>\u2218<\/mo><\/msup>90^{\\circ}<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q9. C. cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi>=<\/mo>b<\/mi>2<\/mn><\/msup>+<\/mo>c<\/mi>2<\/mn><\/msup>\u2212<\/mo>a<\/mi>2<\/mn><\/msup><\/mrow>2<\/mn>b<\/mi>c<\/mi><\/mrow><\/mfrac><\/mrow>\\cos A = \\frac{b^2 + c^2 – a^2}{2bc}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    By moving 2<\/mn>b<\/mi>c<\/mi><\/mspace>cos<\/mi>\u2061<\/mo><\/mspace><\/mrow>A<\/mi><\/mrow>2bc \\cos A<\/annotation><\/semantics><\/math> to the left and a<\/mi>2<\/mn><\/msup>a^2<\/annotation><\/semantics><\/math> to the right, then dividing by 2<\/mn>b<\/mi>c<\/mi><\/mrow>2bc<\/annotation><\/semantics><\/math>, we isolate the cosine of the angle.<\/p>\n\n\n\n

    Q10. B. 9.40<\/mn> m<\/mtext><\/mrow>9.40\\text{ m}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Using h<\/mi>=<\/mo>10<\/mn><\/mspace>sin<\/mi>\u2061<\/mo><\/mspace><\/mrow>70<\/mn>\u2218<\/mo><\/msup>\u2248<\/mo>9.40<\/mn><\/mrow>h = 10 \\sin 70^{\\circ} \\approx 9.40<\/annotation><\/semantics><\/math>. This is a basic right-angled application to contrast with the newer general rules.<\/p>\n\n\n\n

    <\/p>\n","protected":false},"excerpt":{"rendered":"

    To master Unit 2 of General Mathematics, you must move beyond the right-angled trigonometry of earlier years. Chapter 4 introduces the Sine Rule, the Cosine Rule, and the Area Rule, which allow you to solve for any triangle, anywhere. This skill is critical for advanced navigation, surveying, and 3D architectural modeling. Concepts and Skills Covered: […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[],"class_list":["post-980","post","type-post","status-publish","format-standard","hentry","category-math"],"_links":{"self":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/980","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=980"}],"version-history":[{"count":4,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/980\/revisions"}],"predecessor-version":[{"id":996,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/980\/revisions\/996"}],"wp:attachment":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=980"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=980"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=980"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}