Mr Barton’s Maths Notes

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Presentation transcript:

Mr Barton’s Maths Notes Trigonometry 2. Sin, Cos, Tan www.mrbartonmaths.com

2. Sin, Cos and Tan 1. The Crucial Point about Sin, Cos and Tan Just like Pythagoras Theorem, all the work we will be doing with Sin, Cos and Tan only works with RIGHT-ANGLED TRIANGLES So… if you don’t have a right-angled triangle, you might just have to add a line or two to make one! 2. Checking your Calculator is in the Correct Mode Every now and again calculators have a tendency to do stupid things, one of which is slipping into the wrong mode for sin, cos and tan questions, giving you a load of dodgy answers even though you might be doing everything perfectly correctly! Here is the check: Work out: sin 30 And if you get an answer of 0.5, you are good to go! If not, you will need to change into degrees (DEG) mode. Each calculator is different, but here’s how to do this on mine:

3. Labelling the Sides of a Right-Angled Triangle Before you start frantically pressing buttons on your calculator, you must work out which one of the trig ratios (sin, cos or tan) than you need, and to do this you must be able to label the sides of your right-angled triangle correctly. This is the order to do it: 1. Hypotenuse (H) – the longest side, opposite the right-angle 2. Opposite (O) – the side directly opposite the angle you have been given / asked to work out 3. Adjacent (A) – the only side left! O H H H A O A A O Note: is just the Greek letter Theta, and it is used for unknown angles, just like x is often used for unknown lengths!

4. The Two Ways of Solving Trigonometry Problems Both methods start off the same: 1. Label your right-angled triangle 2. Tick which information (lengths of sides, sizes of angles) you have been given 3. Tick which information you have been asked to work out 4. Decide whether the question needs sin, cos or tan The difference comes now, where you actually have to go on and get the answer. Both of the following methods are perfectly fine, just choose the one that suits you best! (a) Use the Formulas and Re-arrange If you are comfortable and confident re-arranging formulas, then this method is for you! Just learn the following formulas: Now just substitute in the two values you do know, and re-arrange the equation to find the value you don’t know!

sin θ h o cos θ h a tan θ a o S O H C A H T O A o = Sin θ x h (b) Use the Formula Triangles This is a clever little way of solving any trig problem. Just make sure you can draw the following triangles from memory: sin θ h o cos θ h a tan θ a o A good way to remember these is to use the initials, reading from left to right: S O H C A H T O A And make up a way of remembering them (a pneumonic, is the posh word!). Now, I know a good one about a horse, but it might be a bit too rude for this website… Anyway, once you have decided whether you need sin, cos or tan, just put your thumb over the thing (angle or side) you are trying to work out, and the triangle will magically tell you exactly what you need to do! Finding Opposite: Finding Hypotenuse: o = Sin θ x h h = a ÷ Cos θ

1. √ √ √ √ o = Tan θ x a 14.3 cm (1dp) Examples ? 500 12 cm H ? O 500 Okay, here we go: 1. Label the sides 2. Tick which information we have been given… which I reckon is the angle and the Adjacent 3. Tick which information we need… which I reckon is the Opposite side 4. Decide whether we need sin, cos or tan … well, looking above, the only one that contains both O and A is… Tan! H √ ? O √ 500 A 12 cm √ 6. And now we know how to do it! 5. Now we place our thumb over the thing we need to work out, which is the Opposite: o = Tan θ x a 14.3 cm (1dp)

2. √ √ √ √ h = a ÷ Cos θ 3.45 m (2dp) 3.1 m ? 260 O 3.1 m H A ? 260 Okay, here we go: 1. Label the sides 2. Tick which information we have been given… which I reckon is the angle and the Adjacent 3. Tick which information we need… which I reckon is the Hypotenuse side 4. Decide whether we need sin, cos or tan … well, looking above, the only one that contains both A and H is… Cos! 3.1 m √ H A ? √ √ 260 6. And now we know how to do it! 5. Now we place our thumb over the thing we need to work out, which is the Hypotenuse: h = a ÷ Cos θ 3.45 m (2dp)

3. √ √ √ √ Sin θ = o ÷ h Sin θ = 0.740740740740… 47.790 (2dp) ? 8 mm H A Okay, here we go: 1. Label the sides 2. Tick which information we have been given… which I reckon is the Hypotenuse and the Opposite 3. Tick which information we need… which I reckon is the angle 4. Decide whether we need sin, cos or tan … well, looking above, the only one that contains both O and H is… Sin! 6 mm O √ 6. Okay, be careful here: Sin θ = o ÷ h 5. Now we place our thumb over the thing we need to work out, which is the angle… or Sin θ Sin θ = 0.740740740740… But that’s not the answer! We don’t want to know what Sin θ is, we want to know what θ is, so we must use “inverse sin” to leave us with just θ on the left hand side: 47.790 (2dp)

4. a = o ÷ Tan θ 3.639702… 7.28 cm (2dp) 10 cm 400 H O ? 10 cm 700 ? A Now, we have a problem here… we don’t have a right-angled triangle! But we can easily make one appear from this isosceles triangle by adding a vertical line down the centre, and then we can carry on as normal… 10 cm 400 H O ? 10 cm 1. Label the sides 2. Tick which information we have been given… which I reckon is the angle and the Opposite 3. Tick which information we need… which I reckon is the Adjacent 4. Decide whether we need sin, cos or tan … well, looking back, the only one that contains both O and A is… Tan! 700 ? A 6. And now we know how to do it! a = o ÷ Tan θ 5. Now we place our thumb over the thing we need to work out, which is the Adjacent 3.639702… But that’s not the answer! We’ve only worked out half of the base of the isosceles triangle! So we need to double this to give us our true answer of: 7.28 cm (2dp)

Good luck with your revision!