The sine rule and the area of a triangle

A triangular sail has angles of forty-eight and sixty-seven degrees and one edge of five metres. None of its corners is a right angle. Pythagoras needs a right angle. Sine, cos and tan, the way you've used them so far, need one too. This sail hasn't got one. One rule handles any triangle at all, and it finds the sail's other edge in a single line.

This is video five of eleven in Pythagoras, Trigonometry and Coordinate Geometry. The sine rule leans on using sine with a calculator, so if that's rusty, revisit Using trigonometry to find missing sides and angles.

This video is Higher tier, or Extended if your course uses Core and Extended.

The first check on any triangle is for a right angle. Opposite, adjacent and hypotenuse, the side facing the right angle, only exist around one. Without a right angle, those three labels have nothing to stand on. A right angle shows up in one of two ways. Either a small square is drawn in the corner, or two angles you know add up to ninety, which leaves ninety for the third. Triangle A has a small square drawn in one corner. B has fifty and sixty degrees, C has thirty and sixty. Which one has no right angle? B. Its third angle is seventy. For C, thirty plus sixty is ninety, so its third angle is a hundred and eighty take away ninety, which is ninety. C is right-angled with no mark at all. Triangle B is scalene, meaning all three sides are different lengths. A scalene triangle with no right angle is exactly where the sine rule comes in. One examiner's report describes what happened on a triangle like that. "The most popular, although incorrect, approach here was to assume triangle ABC was right-angled rather than scalene and to use one of the familiar trigonometry ratios in an attempt to find the length of a side." Those students used a right-angle tool on a triangle without a right angle. The fix is to check for the square, or add up the angles, before any tool is chosen.

The sine rule starts with a labelling habit. Capital letters A, B and C name the angles. Small letters a, b and c name the sides, and each small letter sits opposite its own capital. Every angle faces its own side - look straight across. From angle A you see side a, from B you see b, and from C you see c. In triangle P Q R, which side does angle Q face: p, q or r? Little q, the small letter matching its capital. The sine rule says: a over sine A equals b over sine B equals c over sine C. In words, each side divided by the sine of the angle it faces gives the same number, all the way round the triangle. Here is where it comes from. Drop a line from corner C straight down to side c, meeting it at a right angle, and call its length h. That line splits the triangle into two right-angled halves. In the first half the hypotenuse is b and h faces angle A, so sine A is h divided by b. Multiply by b, and h is b sine A. In the second half the hypotenuse is a, so the matching move makes h a sine B. Both of those equal h. What equation joins b sine A and a sine B? Both describe the same height, so b sine A equals a sine B. Divide both sides by sine A and by sine B, and what's left is a over sine A equals b over sine B. In a question you only use two of the three fractions: one complete pair, a side with its facing angle both known, and one pair that holds the unknown.

The sail's missing edge comes first. Angle A is forty-eight degrees, angle B is sixty-seven degrees, and side a, facing A, is five metres. The side facing B is wanted. Which pair is complete, and which pair holds the unknown? Little a with capital A is complete, with both values known. Little b with capital B holds the unknown. Write those two fractions. b over sine sixty-seven equals five over sine forty-eight. To get b alone, multiply both sides by sine sixty-seven. That gives b equals five times sine sixty-seven, divided by sine forty-eight. Press five, times, sine, sixty-seven, close the bracket, divide, sine, forty-eight, close the bracket, equals. The display reads six point one nine three three zero seven two nine three. To one decimal place, the side facing B is six point two metres. It's longer than the five, which fits, because it faces the bigger angle.

The same rule finds angles. In triangle A B C, angle A is forty-two degrees, side a is seven centimetres and side b is nine centimetres. Angle B is wanted, and then angle C. When an angle is wanted, flip the fractions so the sines are on top: sine A over a equals sine B over b. The unknown then sits on the top, where it's easiest to reach. Put the numbers in. Sine B over nine equals sine forty-two over seven. Multiply both sides by nine. Sine B equals nine times sine forty-two, divided by seven. Press nine, times, sine, forty-two, close the bracket, divide, seven, equals. The display reads nought point eight six zero three one zero seven seven nine six. That's sine B, not B itself, so take the inverse using the full value. Press shift, sine, the answer key, equals. The display reads fifty-nine point three five one four nine five one three. To one decimal place, angle B is fifty-nine point four degrees. The diagram draws angle B as acute, so this is the value we want. Next, angle C: the sine rule again, or something simpler? Something simpler. The three angles in a triangle add up to a hundred and eighty, so C is whatever is left over. A hundred and eighty take away forty-two is a hundred and thirty-eight. Take away B's full value, fifty-nine point three five one five, and that leaves seventy-eight point six four eight five. To one decimal place, angle C is seventy-eight point six degrees. One examiner's report describes a question built in the same two stages. "Question fifteen b was a two-stage sine rule question, in that they had to find angle ABC first and then subtract the two angles from a hundred and eighty to find angle ACB." Reaching for another rule at that second stage only adds working. The fix is to stop once two angles are known: the third is a hundred and eighty take away the other two.

One more formula works on any triangle, and it gives the area. Area equals a half, times a, times b, times sine C. Here a and b are two sides, and C is the angle between them, the corner where they meet. Angle C faces side c, the one side the formula leaves out. It's half base times height in disguise. Take side a as the base. The height up to the third corner is b times sine C, so the area is a half, times a, times b sine C. A triangular flag has angle A of fifty degrees, angle B of sixty-five degrees, and side a of six metres. The task is side b with the sine rule, then the flag's area. b over sine sixty-five equals six over sine fifty. That makes b equal to six times sine sixty-five, divided by sine fifty. Press six, times, sine, sixty-five, close the bracket, divide, sine, fifty, equals. The display reads seven point zero nine eight six zero four seven four nine. That's seven point one metres to one decimal place, but the full value stays in the calculator. Sides a and b are known now. Which angle goes into the area formula, and why? Angle C, because it's the corner between those two sides. It isn't given, so it has to be found first. The angles add up to a hundred and eighty. A hundred and eighty take away fifty, take away sixty-five, leaves sixty-five, so angle C is sixty-five degrees. Now the formula. Area equals a half, times six, times seven point zero nine eight six, times sine sixty-five. Press nought point five, times, six, times the answer key, which holds b's full value, times, sine, sixty-five, equals. The display reads nineteen point three zero zero five six two two eight. To one decimal place, the flag's area is nineteen point three square metres. One examiner's report, on a triangle question that ended in an area, says this. "It is considered to be a problem solving question because the candidates have to make a decision to find an angle and which angle to find." Skip that decision and the formula gets the wrong angle, or no angle at all. The fix is to pick the two sides you'll multiply first, then find the angle where they meet. Every angle faces its own side - look straight across. In the sine rule, each side sits over the sine of the angle it faces; in the area formula, the angle sits between its two sides.

The sail had no right angle and still gave up its edge. A short test to finish, and the rule itself comes first. What is the sine rule, all three fractions? a over sine A equals b over sine B equals c over sine C, each side over the sine of the angle it faces. Now try another: A thirty degrees, a six, B forty-five. How long is b, by the sine rule? About eight point five. b is six times sine forty-five over sine thirty, which is six root two. One more, and this time the answer is an area. Sides of eight and eleven centimetres meet at forty degrees. What's the area? Twenty-eight point three square centimetres. A half times eight times eleven is forty-four, and forty-four times sine forty is twenty-eight point two eight.

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Next in the chapter: Trigonometric ratios of obtuse angles, for Higher.

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