Topic 2: The Trigonometric Ratios Finding Sides

Transcription

1 Topic 2: The Trigonometric Ratios Fining Sies Labelling sies To use the Trigonometric Ratios, commonly calle the Trig Ratios, it is important to learn how to label the right angle triangle. The hypotenuse of the triangle is still the longest sie an locate opposite the right angle. The two other sies are name: the opposite an the ajacent. The everyay language meanings of these terms help in using these labels. The opposite sie is locate opposite the specifie angle. The ajacent is locate ajacent (net to) the specifie angle. Opposite Hypotenuse θ This is the specifie angle. Ajacent When naming a triangle, always name the Hypotenuse first. The reason for oing this is that both the Hypotenuse an the Ajacent are ajacent (net to) the specifie angle. By naming the hypotenuse first there can only be one sie that is ajacent to the specifie angle. Eample: 18 Ajacent Hypotenuse Opposite Ajacent 20 Hypotenuse Opposite Ajacent 55 Hypotenuse Opposite Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 1

2 If there are two angles, the opposite an ajacent are still relative to the specifie angle. The iagram below shows the iea. This is the specifie angle α. α Hypotenuse Opposite the specifie angle θ Ajacent the specifie angle α θ Hypotenuse This is the specifie angle θ. Ajacent the specifie angle θ Opposite the specifie angle α The Trigonometric Ratios With right angle triangles having three sies, it is possible to have 6 ratios. They are: opposite ajacent opposite hypotenuse ajacent hypotenuse,,,, an. ajacent opposite hypotenuse opposite hypotenuse ajacent There are basically 3 unique ratios. The 3 trig ratios commonly use are given the names Sine, Cosine an Tangent. These names are abbreviate to Sin, Cos an Tan. If the specifie angle is θ, then the ratios are written as: opposite Sin θ = hypotenuse ajacent Cos θ = hypotenuse opposite Tan θ = ajacent A challenge for you is to evise a way to remember these ratios. The table below show how the ratios are applie to right angle triangles. a c For the specifie angle θ: opposite a Sin θ = = hypotenuse c ajacent b Cos θ = = hypotenuse c b θ opposite a Tan θ = = ajacent b Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 2

3 12 H 5 For the specifie angle J: 5 Sin J = 13 For the specifie angle I: 12 Sin I = I 12 Cos J = 13 5 Cos I = 13 J 5 Tan J = Tan I = 5 14 m For the specifie angle of 43 Sin 43 = 14 Cos 43 = y y Tan 43 = y To help unerstan trigonometry, the eample below gives an inication of how it works. Consier a right angle triangle with an angle of 45, this means that the missing angle is also 45. This means that the opposite an ajacent sie must be equal, so the triangle is actually an isosceles right angle triangle. The triangle is rawn below. hypotenuse 45 Ajacent 90-45=45 Opposite Same length opposite The Trig ratio associate with opposite an ajacent is Tanθ =. With the opposite an ajacent the ajacent being the same length, the value of this ratio woul be 1. So far: In a right angle triangle with a specifie angle of 45, the value of the ratio opposite is 1. This means Tan 45 is always 1. ajacent Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 3

4 Because the value of the ratio is obtaine without actually knowing any measurements, another key iea is: Key Iea For any size right angle triangle with a specifie angle of 45, opposite the ratio Tan 45 = is always 1. If the specifie angle is 45, the length ajacent of the opposite an the ajacent must be the same. There are other ratios that can be calculate an whatever value they equal, the value will be relate the 45 angle. This iea can be applie to right angle triangles with other sizes of specifie angles. Using your calculator The net step is learning how to use the Trig functions on your scientific calculator. At this stage the trig ratio associate with a given angle will be foun. Scientific calculators allow the user to epress the angle in three ifferent ways. In this moule, angles are being epresse in egrees. Rea the user guie for your calculator to make sure it is always set on egrees. On the Sharp EL 531, there is a key labelle DRG. Pressing this key allows the user to change between the ifferent angle types. Keep pressing until DEG is foun on the upper part of the screen. Note: on this calculator it is easy to change this setting accientally. Every time the user intens to use Trig functions, this setting shoul be checke. On the Casio f-82au, press the SHIFT key then the MODE SETUP key. Press 3 for Degrees. A small D shoul appear on the top of the isplay as shown on the left. Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 4

5 Alreay, it is known from earlier in this topic that Tan 45 = 1. This can be confirme by pressing l45=, the isplay on the calculator shoul rea 1. (If not, check the egrees setting an try again.) If you have an oler style scientific calculator, you may have to o 45l=. Use your calculator to fin the following: sin 30 (your calculator shoul isplay 0.5) This means that in a right angle triangle with a specifie angle of 30, the opposite sie is 0.5 the length of the hypotenuse. cos 72 (your calculator shoul isplay ) This means that in a right angle triangle with a specifie angle of 72, the ajacent sie is (appro) 0.31 the length of the hypotenuse. tan 19 (your calculator shoul isplay ) This means that in a right angle triangle with a specifie angle of 19, the opposite sie is (appro) 0.34 the length of the ajacent. Fining missing sies In this right angle triangle, calculate the length of the sie marke. 37 Step 1: Determine which ratio to use. In this triangle, one angle an one sie are given (37, 12m). Base on the 37 (specifie) angle, the 12m sie is the ajacent an the sie is the opposite. The trig ratio to be use in this question is Tan because it contains the sie lengths ajacent an opposite. Step 2: Write out the ratio an substitute in values. opp Tanθ = aj Tan37 = m Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 5

6 Step 3: Do the necessary rearranging. 12 Tan37 = 12 multiply both sies by Tan37 = Step 4: Calculate the answer using your calculator. 12[l37= The missing sie length () is appro 9.04m. Step 5: Try to check the reasonableness The 12m sie is opposite an angle of (90-37) 53, because is opposite the 37 angle it shoul be smaller than 12m, so 9.04 coul be correct. (small sies are opposite small angles, large sies are opposite large angles) A force of 1500N acts at 36 below the right horizontal as shown in the iagram below. Calculate the size of the vertical component of the force (v) 36 Force = 1500N Vertical Component (V) Step 1: Determine which ratio to use. In this triangle, one angle an one sie are given (36, 1500N). Base on the 36 (specifie) angle, the 1500N sie is the hypotenuse an the v sie is the opposite. The trig ratio to be use in this question is Sin because it contains the sie lengths hypotenuse an opposite. Step 2: Write out the ratio an substitute in values. opp Sinθ = hyp Sin36 = 1500 Step 3: Do the necessary rearranging. v 1500 Sin36 = 1500 multiply both sies by Sin36 = v Step 4: Calculate the answer using your calculator. Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 6

7 1500[j36=881.7 The vertical component of force (v) is appro 881.7N. Step 5: Try to check the reasonableness The hypotenuse of this triangle is 1500N. The hypotenuse is also the longest sie. Obtaining v as N is consistent with this information. The net problem is slightly ifferent in metho. Step 3: rearranging is ifferent because the variable is in the enominator. 455 m 19 Step 1: Determine which ratio to use. In this triangle, one angle an one sie are given (19, 455m). Base on the 19 angle, the 455m sie is the ajacent an the sie is the hypotenuse. The trig ratio to be use in this question is Cos because the question contains the two sie lengths ajacent an the hypotenuse. Step 2: Write out the ratio an substitute in values. aj Cosθ = hyp 455 Cos19 = Step 3: Do the necessary rearranging. 455 Cos19 = multiply both sies by Cos19 Cos = ivie both sies by Cos19 Cos = Cos19 Step 4: Calculate the answer using your calculator. 455Pk19= The missing sie length is appro 481.2m Step 5: Try to check the reasonableness The missing sie is the hypotenuse which is the longest sie on the triangle. The answer of 481.2m is longer than the other given sie (455m) so the answer coul be correct. Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 7

8 Some problem solving questions refer to the Angle of Elevation or the Angle of Depression The angle of elevation is the angle forme by a line of sight above the horizontal Line of sight Angle of elevation Horizontal Horizontal The angle of epression is the angle forme by a line of sight below the horizontal Angle of epression Line of sight The net problem is a simple problem solving question. The etra skill here is to rea the information given an construct a iagram. The angle of elevation of a plane at an altitue of 4500m is 27 to the horizontal. In a irect line, how far away is the plane. 4500m 27 Step 1: Determine which ratio to use. In this triangle, one angle an one sie are given (27, 4500m). Base on the 27 angle, the 4550m sie is the opposite an the sie is the hypotenuse. The trig ratio to be use in this question is Sin. Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 8

9 Step 2: Write out the ratio an substitute in values. opp Sinθ = hyp 4500 Sin27 = Step 3: Do the necessary rearranging Sin27 = multiply both sies by Sin27 Sin = ivie both sies by Sin27 Sin = Sin27 Step 4: Calculate the answer using your calculator. 4500Pj27= The plane is 9912m away. Step 5: Try to check the reasonableness The missing sie is the hypotenuse, which is the longest sie on the triangle. The answer of 9912m is longer than the other given sie (4500m) so the answer coul be correct. Vieo Trigonometry Ratios Fining Sie Lengths The sections below may or may not be relevant to your stuies. Activity questions are locate at the en. Angles less than 1 or containing a part angle Angles with a part that is smaller than 1 egree can be epresse either in egrees as a ecimal or in egrees, minutes an secons. Angle measurement in egrees, minutes an secons is just like time (hours, minutes an secons). 60 secons equalling 1 minute (60 = 1 ) 60 minutes equalling one egree (60 = 1 ) Fining the tan of 32.7 is l32.7=, the isplay shoul rea Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 9

10 Fining the sin of (67 egrees, 42 minutes) is: On a Casio f-82: j6742= The key represents egrees( ), minutes( ) an secons( ) Or On a Sharp EL531: j67 The isplay shoul rea D M ' S 42= Fining the cos of (37 egrees, 22 minutes, 41 secons ) is: On a Casio f-82: k Or On a Sharp EL531: k37 D M ' S 22 D M ' S 41 The isplay shoul rea Use your calculator to fin the following: sin 14.5 (your calculator shoul isplay ) cos (your calculator shoul isplay ) tan 27.1 (your calculator shoul isplay ) sin (your calculator shoul isplay ) Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 10

11 Compass an True Bearings This problem involves compass irections. There are two methos for epressing irections. The first metho is a True Bearing where North is 0 an the angle increases in a clockwise irection, giving East as 90, South as 180 an West as 270. N 40 T = 303 T = 239 T The secon metho is a Compass Bearing such as S40 W. The first compass irection state is either N or S, followe by an angular measurement in an E or W irection. N N 57 W N 40 E 31 S 59 W Vieo Bearings A group of bushwalkers walk 5.4 km north an then turn. They walk in an easterly irection until they reach a tower. The bearing of the tower from the original point is N44 E. Calculate the istance walke in the easterly irection by the walkers. Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 11

12 The iagram for the problem is rawn below. Tower 1. Determine which ratio to use. In this triangle, one angle an one sie are given (44, 5.4km). Base on the 44 angle, the 5.4km sie is the ajacent an the sie is the opposite. The trig ratio to be use in this question is Tan. 5.4 km Write out the ratio an substitute in values. opp Tanθ = aj Tan44 = Do the necessary rearranging. 5.4 Tan44 = Tan44 = 4. Calculate the answer using your calculator. 5.4Ol44= The tower is 5.2km to the east. 5. Try to check the reasonableness The missing sie shoul be about the same length as the given sie. The net eample is a problem solving question that contains multiple steps. Care must be taken to rea the question an accurately transfer this information into a iagram. A ship is 1km out to sea from the base of a cliff. On top of the cliff is a lighthouse. From the ship, the angle of elevation to the base of the lighthouse is 16 an the angle of elevation to the top of the lighthouse is Calculate the height of the lighthouse. The iagram is; h L h t h c km Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 12

13 Remember that only right angle triangles can be use to solve this question. The strategy to solve this problem is: 1. Calculate the height of the cliff 2. Calculate the total height (cliff + lighthouse) 3. Calculate the height of the lighthouse by subtracting the height of the cliff from the total height. Calculating the height of the cliff: (1km = 1000m) hc Tan16 = Tan16 = h Calculating the total height: = h ht Tan19.5 = Tan19.5 = h = ht Calculating the height of the lighthouse: h = h h h c L t c L = hl = 67.4m The height of the lighthouse is 67.4 m. This is approimately equivalent to a 14 level builing! c t Activity 1. For the following triangles, ientify the ajacent, opposite an hypotenuse for each triangle. (a) (b) (c) a α z y b c 60 Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 13

14 5 θ () (e) (f) 45.34mm 245 β γ 30mm 34mm 25 h 450 k δ g 2. For the following triangles, fin the value of the missing sie(s). (a) (b) (c) m h f b 58 Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 14

15 () (e) (f) 14.7m b e m cm 24 45' m (g) (h) (i) ' 345mm h h 62.5 km cm 57 Fin h an then the area of the triangle. 3. Epress the following irections as a true bearing. (a) S15 E (b) NE (c) N45 W () 10 W of N (e) SSW (f) A person staning on top of a 25 m cliff sees a rower out to sea. The angle of epression of the boat is How far is the boat from the base of the cliff? 5. A stuent wishes to fin the height of a builing. From a istance of 50m on perfectly level groun, the angle of elevation to the top is Fin the height of the builing? 6. A plane is flying at altitue of 5000m. The pilot observes a boat at an angle of epression of 12, calculate the horizontal istance which places the plane irectly above the boat. 7. A walker ecies to take a irect route to a lanmark. They walk 1.7 km at a bearing of 78 T. How far i they walk in a northerly an easterly irection? Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 15

16 8. Fin the perimeter of this trapezium. 28 cm cm 9. A kite is attache to a 45m line. On a winy ay, the kite flies at an angle of elevation of 28. Calculate the height of the kite above the groun. 10. A plane flying at an altitue of m is flying away from a person. The angle of elevation of the plane is 76 when initially observe. After 1 minute 15 secons, the plane is at an angle of elevation of 29. Ignoring the height of the person, what is the spee of the plane in km/hr? Centre for Teaching an Learning Acaemic Practice Acaemic Skills Page 16