Trigonometry in any triangle · Grade 10 · 5 units
In Trigonometry in any triangle (Grade 10 · 5 units) you practise 7 skills: The area of a triangle with the sine · The sine rule · The cosine rule · One triangle or two · Angles with the same sine · The largest value without a derivative · Solving a triangle in two steps. The area of any triangle, not only a right triangle, is half the product of two sides times the sine of the angle between them. For an obtuse angle, use the fact that an angle and the angle that makes it up to 180 degrees have the same sine. So the same area can come from two different angles, an acute one and an obtuse one.
The sine rule says that in every triangle a side over the sine of the angle opposite it is the same for all three sides, and it equals the diameter of the circle through the corners. The cosine rule finds a side from the other two sides and the angle between them, or an angle from the three sides. When two sides and an angle that is not between them are given, there can be two triangles, one or none. When the question asks for an exact answer, keep the roots: for example, the sine of 60 degrees is root three over two. When it asks for rounding, use a calculator and round only at the end. A sine is never more than one, so you can find a largest value without a derivative. A model proof of the sine rule: write the area of the triangle in three ways, each time with two sides and the sine of the angle between them. The three expressions are equal. Divide each of them by half the product of all three sides, and what is left is the sine of each angle over the side facing it, the same number for all three angles. Turn the fractions over and you have the sine rule. The cosine rule is proved with the altitude from one corner: Pythagoras’ theorem in the two right triangles the altitude makes, and the cosine for the part of the base next to the angle.
Tip: Before you calculate, mark what is given on the triangle: two sides and the angle between them lead to the cosine rule, and a side with the angle opposite it leads to the sine rule.
What you learn
The area of a triangle with the sine
For example: In triangle ABC , in centimetres, AB = 4, AC = 12, ∢A = 30° . What is the area of the triangle?
Answer and solution
The answer: 12 cm²
The sine of 30° is 1/2 , so S = 4 × 12 × 1/2 ÷ 2 = 12 cm².
The sine rule
For example: In triangle ABC you are told that AC = 6, ∢B = 30°, ∢C = 100° . How long is the side AB in cm? Give your answer to 2 decimal places.
Answer and solution
The answer: 11.82 cm
The side AB is opposite the angle at C , and the side AC is opposite the angle at B : AB / sin 100° = 6 / sin 30° , so AB = 6 × sin 100° / sin 30° ≈ 11.82 cm.
The cosine rule
For example: In triangle ABC , in centimetres, AB = 3, AC = 8, ∢A = 60° . How long is BC ?
Answer and solution
The answer: 7 cm
The cosine of 60° is 1/2 , so BC² = 3² + 8² − 2 × 3 × 8 × 1/2 = 9 + 64 − 24 = 49 . So BC = 7 cm.
One triangle or two
For example: In triangle ABC , in centimetres, AC = 6, BC = 3, ∢A = 70° . How many such triangles are there?
- One triangle
- Two triangles
- There is no such triangle
Answer and solution
The answer: There is no such triangle
By the sine rule, sin B = AC × sin A / BC = 6 × sin 70° / 3 ≈ 1.8794 . A sine is never more than 1, so there is no such triangle.
Angles with the same sine
For example: You are told that sin α = sin 20° , and the angle α is obtuse. What is α ?
Answer and solution
The answer: 160 °
The angle 20° is acute, and the angle that makes it up to 180° has the same sine: α = 180° − 20° = 160° .
The largest value without a derivative
For example: What is the largest value of 4 sin 2α ?
Answer and solution
The answer: 4
A sine is never more than 1, so 4 sin 2α ≤ 4 . It is equal when 2α = 90° , that is when α = 45° . The largest value is 4 .
Solving a triangle in two steps
For example: In triangle ABC , in centimetres, AB = 5, AC = 12, BC = 9 . What is the area of the triangle? Give your answer to 2 decimal places.
Answer and solution
The answer: 20.4 cm²
By the cosine rule, cos A = (5² + 12² − 9²) / (2 × 5 × 12) ≈ 0.7333 , so sin A = √(1 − cos²A) ≈ 0.6799 . The area is S = 5 × 12 × sin A ÷ 2 ≈ 20.4 cm².
Not on the site yet
Parts of this topic that are not built yet: Proving the area formula, the sine rule and the cosine rule · Geometric proofs with trigonometry.
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 5 units (grades 10 to 12).