Sine, cosine and area rule grade 11 pdf questions and answers

sine, cosine and area rule grade 11 pdf questions and answers

Sine, Cosine, and Area Rule Grade 11 PDF Questions and Answers

Answer:
Here, I’ll provide a detailed understanding and examples of sine, cosine, and the area rule, often taught in Grade 11 trigonometry. This explanation will cover fundamental principles, formulas, and solutions to sample problems, emulating a high-quality educational resource.

Sine Rule

The Sine Rule (also known as the Law of Sines) is used in non-right angled triangles to relate the lengths of the sides to the sines of their respective opposite angles. The rule states:

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Cosine Rule

The Cosine Rule (or Law of Cosines) extends the Pythagorean theorem to all triangles, making it possible to calculate an unknown side or angle. The formula is:

For sides:

c^2 = a^2 + b^2 - 2ab \cos C
b^2 = a^2 + c^2 - 2ac \cos B
a^2 = b^2 + c^2 - 2bc \cos A

For angles:

\cos A = \frac{b^2 + c^2 - a^2}{2bc}
\cos B = \frac{a^2 + c^2 - b^2}{2ac}
\cos C = \frac{a^2 + b^2 - c^2}{2ab}

Area Rule

The area of any triangle can be found using the formula:

\text{Area} = \frac{1}{2}ab \sin C

Step-by-Step Solutions to Example Problems

Problem 1: Using the Sine Rule

Q: In triangle ABC, where A = 30^\circ, B = 45^\circ, and a = 10 cm, find the length of side b.

Solution:

  1. Apply the Sine Rule:

    \frac{a}{\sin A} = \frac{b}{\sin B}
  2. Insert known values:

    \frac{10}{\sin 30^\circ} = \frac{b}{\sin 45^\circ}
  3. Simplify and solve for b:

    \frac{10}{0.5} = \frac{b}{0.7071}
    20 = \frac{b}{0.7071}
    b = 20 \times 0.7071 \approx 14.14 \, \text{cm}

Final Answer:
b \approx 14.14 \, \text{cm}

Problem 2: Using the Cosine Rule

Q: In triangle XYZ, x = 7 cm, y = 5 cm, and \angle Z = 60^\circ. Find the length of side z.

Solution:

  1. Apply the Cosine Rule:

    z^2 = x^2 + y^2 - 2xy \cos Z
  2. Insert known values:

    z^2 = 7^2 + 5^2 - 2 \cdot 7 \cdot 5 \cdot \cos 60^\circ
  3. Simplify:

    z^2 = 49 + 25 - 2 \cdot 7 \cdot 5 \cdot 0.5
    z^2 = 49 + 25 - 35
    z^2 = 39
    z = \sqrt{39} \approx 6.24 \, \text{cm}

Final Answer:
z \approx 6.24 \, \text{cm}

Problem 3: Finding the Area of a Triangle Using the Area Rule

Q: In triangle PQR, p = 8 cm, q = 6 cm, and \angle R = 45^\circ. Find the area of triangle PQR.

Solution:

  1. Apply the Area Rule:

    \text{Area} = \frac{1}{2}pq \sin R
  2. Insert known values:

    \text{Area} = \frac{1}{2} \cdot 8 \cdot 6 \cdot \sin 45^\circ
  3. Simplify:

    \text{Area} = \frac{1}{2} \cdot 8 \cdot 6 \cdot 0.7071
    \text{Area} = 24 \cdot 0.7071 \approx 16.97 \, \text{cm}^2

Final Answer:
Area \approx 16.97 \, \text{cm}^2


By mastering the sine, cosine, and area rules, students can tackle various trigonometric problems involving non-right angled triangles. This guide covers the basic formulas and provides step-by-step solutions to common types of problems, serving as a comprehensive resource for Grade 11 trigonometry students.