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:
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:
For angles:
Area Rule
The area of any triangle can be found using the formula:
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:
-
Apply the Sine Rule:
\frac{a}{\sin A} = \frac{b}{\sin B} -
Insert known values:
\frac{10}{\sin 30^\circ} = \frac{b}{\sin 45^\circ} -
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:
-
Apply the Cosine Rule:
z^2 = x^2 + y^2 - 2xy \cos Z -
Insert known values:
z^2 = 7^2 + 5^2 - 2 \cdot 7 \cdot 5 \cdot \cos 60^\circ -
Simplify:
z^2 = 49 + 25 - 2 \cdot 7 \cdot 5 \cdot 0.5z^2 = 49 + 25 - 35z^2 = 39z = \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:
-
Apply the Area Rule:
\text{Area} = \frac{1}{2}pq \sin R -
Insert known values:
\text{Area} = \frac{1}{2} \cdot 8 \cdot 6 \cdot \sin 45^\circ -
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.