0580 · 6.5Non-right-angled triangles

Area formula and triangle notation

0:00Imagine measuring a triangular piece of land when none of its corners is a right angle. The familiar right-angled triangle methods cannot be applied directly to the whole shape. We need relationships that work for any triangle. The sine rule, cosine rule and triangle area formula give us those relationships. By the end, you will understand which rule to choose and why it works. You will also see why some measurements can produce two different triangles.

0:24We will use triangle PQR throughout. Each lowercase side letter names the side opposite the matching capital-letter vertex. Side p is opposite angle P, side q is opposite angle Q, and side r is opposite angle R. The formula sheet writes the same rules using the letters a, b and c. Changing the letters does not change the mathematics. Matching each side with its opposite angle is what matters. You need three starting ideas from earlier work.

0:54The angles of a triangle add up to one hundred and eighty degrees, a base and perpendicular height give an area, and trigonometric ratios work inside right-angled triangles. An obtuse angle is greater than ninety degrees but less than one hundred and eighty degrees. Keep your calculator in degree mode for these calculations. A radian setting would give incorrect answers for angles measured in degrees. Start with the area formula you already know.

1:18Triangle area equals one half multiplied by base multiplied by perpendicular height. Here, choose side r as the base and call the height h. Dropping a perpendicular from R creates a smaller right-angled triangle inside the original shape. In that smaller triangle, the sine of angle P = h ÷ q. Rearranging gives h = q multiplied by the sine of angle P. Substitute that height into the familiar area formula.

1:46The area becomes ½ × q × r multiplied by the sine of angle P. The formula uses two sides and their included angle. Included means the angle between those two sides, at the point where they meet. So ½ × p × q multiplied by the sine of angle R is another version. Angle R lies between sides p and q. An obtuse angle does not break the formula.

2:11The perpendicular height may fall outside the triangle, but it still measures the distance to the base line. An angle and its supplement have the same sine. Here, one hundred and twenty degrees and sixty degrees therefore produce the same positive height factor. That relationship explains why the area remains positive for an obtuse included angle. It also prepares us for an important complication when finding angles. Sine can have the same value for two different angles inside a triangle.

2:39We must sometimes decide whether an acute angle, an obtuse angle, or both are possible.

Sine rule and finding sides

2:43The sine rule follows from calculating the same area in different ways. Equate the expressions using angle P and angle Q, then cancel the common factors. This gives the sine of angle P ÷ p equal to the sine of angle Q ÷ q. Taking reciprocals gives the familiar side-over-sine form. The sine rule says that each side divided by the sine of its opposite angle has the same value.

3:07All three pairs describe the same triangle, so their ratios agree. For calculations, use whichever two pairs contain the information you need. There is no need to write all three every time. Choose the sine rule when you have a complete opposite side and angle pair. You also need another side or angle that can be paired with the unknown. With two known angles, first find the third if necessary.

3:29This gives you the matching pairs needed for the calculation. Our first exam-style example asks us to calculate q. Angle P is thirty degrees, angle Q is forty-five degrees, and p is eight centimetres. We know the complete pair consisting of p and angle P. The unknown q is opposite the other given angle, so the sine rule fits. Write q divided by the sine of forty-five degrees equals eight divided by the sine of thirty degrees.

3:57Multiply both sides by the sine of forty-five degrees to isolate q. Substitute the exact values, √2/2 and one half. Simplifying gives q equal to 8√2 centimetres. That exact answer can be found without a calculator. As a decimal, it is eleven point three centimetres to three significant figures. The answer makes sense because the larger angle has the longer opposite side. Angle Q is larger than angle P, so q must be longer than eight centimetres.

Ambiguous sine rule case

4:25Now suppose angle P is thirty degrees, p is seven centimetres, and q is ten centimetres. This time, calculate the possible values of angle Q. We still have a complete opposite pair, so use the sine rule. However, finding an angle requires an extra check that finding a side did not. Write the sine of angle Q ÷ 10 equals the sine of thirty degrees divided by seven. Rearranging gives the sine of angle Q equal to 5/7.

4:52The inverse sine of 5/7 gives approximately forty-five point five eight five degrees. Keep the full calculator value for the remaining steps. Subtract that angle from one hundred and eighty degrees to obtain approximately one hundred and thirty-four point four one five degrees. For the acute option, subtract thirty degrees and the first angle from one hundred and eighty degrees. The remaining angle is approximately one hundred and four point four one five degrees.

5:18It is positive, so this triangle is possible. For the obtuse option, subtract thirty degrees and the second angle from one hundred and eighty degrees. This gives approximately fifteen point five eight five degrees. That remaining angle is also positive, so the second triangle is possible. Angle Q can therefore be forty-five point six degrees or one hundred and thirty-four point four degrees. This is called the ambiguous case of the sine rule.

5:43Two sides and an angle opposite one of them do not always determine one unique triangle. In the diagram, the fixed-length side can meet the base ray at two positions. Each position creates a triangle satisfying the same measurements. Do not automatically keep both angles whenever you use inverse sine. Check that each candidate leaves a positive third angle, and use any additional information in the question. A sine value greater than one cannot describe a real angle.

6:10Depending on the measurements, there can be no triangle, one triangle, or two triangles.

Cosine rule and finding sides

6:15The cosine rule is useful when no complete opposite side and angle pair is available. Use it for two sides and their included angle, or for all three sides. It connects a side to the other two sides and the angle between them. Its correction term accounts for how widely those sides open. To see where it comes from, split an acute triangle with a perpendicular height. The horizontal part, which we call x, equals q multiplied by the cosine of angle P.

6:39Pythagoras in the left small triangle gives x² + h² equal to q². In the right small triangle, p² = r − x, all squared, plus h². Expand the squared difference to get r squared minus two r times x, plus x², plus h². Replace x² + h² with q². Then substitute q multiplied by the cosine of angle P for x. The result is p squared equals q squared plus r squared minus two q times r multiplied by the cosine of angle P.

7:12At ninety degrees, cosine is zero, so the correction disappears and we recover Pythagoras. For an acute angle, the cosine is positive and the correction is subtracted. For an obtuse angle, cosine is negative, so subtracting that correction increases the result. The opposite side becomes longer as the included angle opens wider. In our next example, q is seven centimetres, r is ten centimetres, and angle P is one hundred and twenty degrees.

7:40Calculate p using the cosine rule because the given angle lies between the known sides. First calculate p², not p. The final square root converts the squared length into the required length. Substitute the measurements, noting that 2 × 7 × 10 is one hundred and forty. Then p² = 7² + 10² − 140 times the cosine of one hundred and twenty degrees. The cosine value is negative one half.

8:06This gives forty-nine plus one hundred minus one hundred and forty multiplied by negative one half. Therefore, p² = 149 + 70, giving two hundred and nineteen. Taking the square root gives p equal to √219 centimetres. The decimal answer is fourteen point eight centimetres to three significant figures. You can reach the exact square-root answer without a calculator using the known cosine value. Notice that subtracting a negative term increased the squared length.

Finding angles and triangle area

8:37The cosine rule also finds an angle when all three sides are known. Suppose p is thirteen centimetres, q is seven centimetres, and r is eight centimetres. Rearrange the rule to make the cosine of angle P the subject. Its numerator is q² + r² − p². The denominator is 2 × q × r. Substitution gives 49 + 64 − 169, divided by one hundred and twelve. This simplifies to −56/112, or negative one half.

9:10Taking inverse cosine gives angle P equal to one hundred and twenty degrees. A negative cosine tells us that the angle is obtuse. Unlike sine, cosine gives one possible interior angle for each valid value. All three sides determine the triangle’s angles uniquely. Here, the longest side is opposite the largest angle, which provides another useful check. For a final exam-style question, calculate the area when p is eight centimetres, q is eleven centimetres, and angle R is thirty-five degrees.

9:40The angle is between the two given sides, so use the area formula directly. No third side is needed. Begin with ½ × p × q multiplied by the sine of angle R. Substitution gives ½ × 8 × 11 multiplied by the sine of thirty-five degrees. Simplify the numerical factor to forty-four. Evaluating gives approximately twenty-five point two three seven three square centimetres. Round only the final answer to obtain twenty-five point two square centimetres.

Common mistakes and recap

10:12Common mistakes include mismatching opposite pairs, using a non-included angle in the area formula, and forgetting the cosine rule’s final square root. Label the diagram before substituting to prevent these errors. When finding an angle with sine, check its supplement before deciding on the answer. For area, always use square units rather than length units. Paper Two is non-calculator, so practise rearranging these rules and using exact trigonometric values without a calculator.

10:39Paper Four allows calculator evaluation, including inverse trigonometric functions. Unless instructed otherwise, give lengths and areas to three significant figures and angles to one decimal place. Keep unrounded intermediate values, and follow any specific accuracy instruction in the question. The central idea is that these rules connect the same triangle’s sides, angles and height. The sine rule compares opposite pairs, while the cosine rule links three sides with an included angle.

11:06The area formula replaces perpendicular height with a side multiplied by a sine. Choose the relationship that matches the information, then check that your answer describes a possible triangle.