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This lesson covers the sine rule, cosine rule, the area formula using sine, and an introduction to vectors for the Higher tier of AQA GCSE Mathematics. These tools allow you to work with non-right-angled triangles and are essential for the most challenging geometry questions on the paper.
graph TD
A[Non-right-angled triangle] --> B{What do you know?}
B -->|Two angles and a side| C[Sine Rule]
B -->|Two sides and the angle between them| D[Cosine Rule to find a side]
B -->|All three sides| E[Cosine Rule to find an angle]
B -->|Two sides and the included angle for area| F[Area = half ab sin C]
B -->|Two sides and a non-included angle| G[Sine Rule but check for ambiguous case]
Exam Tip: The most common mistake is using SOHCAHTOA on a non-right-angled triangle. SOHCAHTOA only works for right-angled triangles. For all other triangles, use the sine rule or cosine rule.
The sine rule relates sides and their opposite angles in any triangle.
a / sin A = b / sin B = c / sin C
Or equivalently (when finding an angle):
sin A / a = sin B / b = sin C / c
In triangle ABC, angle A = 40 degrees, angle B = 75 degrees, and side a = 10 cm. Find side b.
Solution: a / sin A = b / sin B 10 / sin(40) = b / sin(75) b = 10 x sin(75) / sin(40) b = 10 x 0.9659 / 0.6428 b = 15.03 cm (2 d.p.)
In triangle PQR, side p = 8 cm, side q = 11 cm, and angle P = 35 degrees. Find angle Q.
Solution: sin P / p = sin Q / q sin(35) / 8 = sin Q / 11 sin Q = 11 x sin(35) / 8 sin Q = 11 x 0.5736 / 8 sin Q = 0.7886 Q = inverse sin(0.7886) Q = 52.1 degrees (1 d.p.)
Exam Tip: When using the sine rule to find an angle, remember that sin(theta) can give two possible angles (theta and 180 - theta). Check whether the obtuse angle makes sense in context. This is the "ambiguous case."
The cosine rule is used when you have:
a squared = b squared + c squared - 2bc cos A
In triangle ABC, b = 7 cm, c = 9 cm, and angle A = 52 degrees. Find side a.
Solution: a squared = 7 squared + 9 squared - 2(7)(9) cos(52) a squared = 49 + 81 - 126 x cos(52) a squared = 130 - 126 x 0.6157 a squared = 130 - 77.58 a squared = 52.42 a = square root of 52.42 a = 7.24 cm (2 d.p.)
Rearrange to: cos A = (b squared + c squared - a squared) / (2bc)
In triangle ABC, a = 8 cm, b = 6 cm, c = 10 cm. Find angle A.
Solution: cos A = (6 squared + 10 squared - 8 squared) / (2 x 6 x 10) cos A = (36 + 100 - 64) / 120 cos A = 72 / 120 cos A = 0.6 A = inverse cos(0.6) A = 53.1 degrees (1 d.p.)
Exam Tip: The cosine rule for finding an angle has a minus sign in the numerator: b squared + c squared MINUS a squared. Make sure the side you subtract is opposite the angle you are finding.
For any triangle where you know two sides and the included angle:
Area = (1/2) x a x b x sin C
where a and b are two sides and C is the angle between them.
Find the area of a triangle with sides 8 cm and 11 cm and an included angle of 64 degrees.
Solution: Area = (1/2) x 8 x 11 x sin(64) Area = (1/2) x 88 x 0.8988 Area = (1/2) x 79.09 Area = 39.5 cm squared (1 d.p.)
A vector has both magnitude (size) and direction. Vectors are used to describe translations and to solve geometric proof problems.
Vectors can be written as:
| Operation | Rule |
|---|---|
| Addition | a + b: travel along a then along b |
| Subtraction | a - b = a + (-b): reverse b then add |
| Scalar multiplication | 2a means the same direction, twice the magnitude |
| Negative | -a means the same magnitude, opposite direction |
If a = (3 on top, 2 on bottom) and b = (1 on top, -4 on bottom), find:
(a) a + b = (3+1 on top, 2+(-4) on bottom) = (4 on top, -2 on bottom)
(b) a - b = (3-1 on top, 2-(-4) on bottom) = (2 on top, 6 on bottom)
(c) 3a = (3x3 on top, 3x2 on bottom) = (9 on top, 6 on bottom)
In vector geometry problems, you are typically given vectors for certain sides and asked to find other vectors or prove geometric properties.
In triangle OAB, vector OA = a and vector OB = b. M is the midpoint of AB.
Find vector OM.
Solution: AB = AO + OB = -a + b = b - a AM = (1/2)AB = (1/2)(b - a) OM = OA + AM = a + (1/2)(b - a) = a + (1/2)b - (1/2)a = (1/2)a + (1/2)b
Three points are collinear (lie on the same straight line) if you can show that one vector is a scalar multiple of another.
Show that points P, Q and R are collinear given that PQ = 2a + 3b and PR = 6a + 9b.
Solution: PR = 6a + 9b = 3(2a + 3b) = 3 x PQ
Since PR is a scalar multiple of PQ, the vectors are parallel and they share the common point P.
Therefore P, Q and R are collinear, with R lying on the line through P and Q.
Exam Tip: To prove collinearity, show that one vector is a scalar multiple of the other AND that they share a common point. Both parts are needed for the full proof.
Two lines are parallel if their direction vectors are scalar multiples of each other.
In quadrilateral OABC, OA = a, OB = a + b, OC = b. Show that AB is parallel to OC.
Solution: AB = AO + OB = -a + (a + b) = b OC = b
Since AB = OC = b, AB and OC are parallel and equal in length. Therefore OABC is a parallelogram.
The magnitude (length) of a column vector (x on top, y on bottom) is:
|v| = square root of (x squared + y squared)
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