Using trigonometry to find missing sides and angles

A drone hangs on a tight tether. The line is eighteen metres long, and it makes an angle of fifty-two degrees with the ground. You want the drone's height. Pythagoras can't help here, because Pythagoras needs two sides, and this gives you one side and one angle. One side and one angle is the signal for trigonometry. By the end of this video, the drone's height comes out of one calculator line, and so does an angle nobody gave you.

This is video three of eleven in Pythagoras, Trigonometry and Coordinate Geometry. Everything here leans on labelling the sides, and Trigonometric ratios, labelling sides and choosing sin, cos or tan is where that is taught.

Here is the signal in full. In a right-angled triangle, if an angle is given or an angle is wanted, the tool is trigonometry. If no angle appears anywhere, it's Pythagoras. Back to the drone. The tether faces the square corner, so the tether is the hypotenuse, the longest side. The height faces the fifty-two degrees, so the height is opposite. Sine, cos or tan: which one uses the tether and the height? Sine. Cos and tan both need the ground side, and nobody gave us that. Sine is opposite over hypotenuse. Call the height h, and the equation is sine fifty-two equals h over eighteen. To get h on its own, multiply both sides by eighteen. That gives h equals eighteen times sine fifty-two. With the calculator in degree mode, press eighteen, then sine, then fifty-two, then equals. The display reads fourteen point one eight four one nine three five six. To one decimal place, that's fourteen point two. The drone is fourteen point two metres up, shorter than the eighteen-metre tether, as a side opposite has to be. One examiner's report says this. "Many students did not recognise the need to use trigonometry in this two mark question, with many trying to work with the area, finding missing angles or using Pythagoras' theorem." They reached for area or Pythagoras with an angle sitting in the question. The fix is to read for an angle before you write anything: given or wanted, it means trigonometry.

This time the angle is the unknown. A skateboard ramp is six metres long, and it rises to a height of two point five metres. The ramp and its height form two sides of a right-angled triangle, with the angle wanted at the bottom. Which ratio links the ramp and the height, and why not tan? The ratio is sine. The ramp faces the square corner, so it's the hypotenuse, and the height faces the angle, so it's opposite. Tan would need the ground, which the question never gives. Call the angle theta. Sine theta equals two point five over six. Now the angle is trapped inside the sine. To get it out, you use the inverse, written sine to the minus one, which answers the question of which angle has this sine. Press shift, then sine, to get sine to the minus one. Type two point five, divide, six, close the bracket, then equals. The display reads twenty-four point six two four three one eight three five. To one decimal place, the ramp makes an angle of twenty-four point six degrees with the ground. That gives you the split for the whole topic. To find a side, use sine, cos or tan of the angle. To find an angle, use the inverse, the minus one button. Next: the ramp's base, from the six and the two point five. Which tool, and why? Pythagoras. Two sides are known and a third is wanted, and it needs no angle, so it avoids leaning on a rounded one. The ramp is the hypotenuse, so we subtract. Six squared is thirty-six. Two point five squared is six point two five. Thirty-six take away six point two five is twenty-nine point seven five. The square root of twenty-nine point seven five is five point four five, to two decimal places. The base runs about five point four five metres along the ground. Another examiner's report records the slip that comes after spotting trigonometry. "many recognised this was trigonometry and could quote the three formulae but chose cosine or, less often, tangent." Knowing all three ratios is no help if the wrong one is picked. The fix is a fixed order: label the sides, name the two you're using, then choose the only ratio that holds both.

The last question hands you the answer and asks for the proof. An isosceles triangle, meaning two sides the same length, has equal sides of ten centimetres and a base of twelve centimetres. The task is to show that the angle at the top, between the two equal sides, is about seventy-three point seven degrees. A show-that question gives you the destination. The working has to start from the question's own facts and arrive there. This triangle has no right angle, so we make one. A line from the top straight down to the middle of the base splits it into two right-angled halves. Each half has a hypotenuse of ten centimetres, one of the equal sides, and a bottom side of six centimetres, half of the twelve. The angle at the top is cut in half too. From the half-angle at the top, which ratio links the six and the ten? Sine, because the six faces that half-angle, so it's opposite, and the ten is the longest side. Cos and tan would both need the height of the half, and nobody has given it. Call the half-angle x. Sine x equals six over ten, which is nought point six. Press shift, sine, nought point six, equals. The display reads thirty-six point eight six nine eight nine seven six five degrees. That is only half the top angle, so double it with the full value still showing. Press times two, equals, and the display reads seventy-three point seven three nine seven nine five two nine. To one decimal place, that's seventy-three point seven degrees, the value we were asked to show, reached from the ten and the twelve alone. Here's a version that runs the other way. Half of seventy-three point seven is thirty-six point eight five, and the sine of that is about nought point six, which matches six over ten. What's wrong with working it that way round? It uses the very angle it was meant to prove, so it proves nothing, however neat it looks. One examiner's report on a show-that question puts it plainly. "Candidates need to realise that when the question states 'Show that' they cannot start with the value given." The fix is to start every show-that from the question's own facts, and let the target value appear only on the last line. Every angle faces its own side - look straight across. From the angle you're using, the side you see is opposite, and that one label picks the ratio.

The drone's height took one signal, one ratio and one calculator line. Three checks now, on triangles this video never drew. A question gives two sides and wants the third, with no angle anywhere. Which tool? It is Pythagoras. No angle given or wanted means no trigonometry. Try the next one. Hypotenuse twenty metres, angle thirty-five. How long is the adjacent? Sixteen point four metres. Cos uses adjacent and hypotenuse, and twenty times cos thirty-five is sixteen point three eight, which rounds to sixteen point four. Now an angle: opposite seven, adjacent four. Which button, and what angle? Tan to the minus one. Seven over four is one point seven five, and the inverse tan of that is sixty point three degrees, to one decimal place.

Once this one feels like yours to own, a thumbs-up marks it, so you can skip it when you revise. If the inverse button is still a blur, save the video and give it a day. Look again tomorrow and it usually settles.

Next in the chapter: Angles of elevation and depression.

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