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## Gradients

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**Gradient of a Road**The gradient shows the steepness of the roads. On some inclined roads, we may find road signs with ratios (e.g. 1 : 6) showing the gradients of the roads.**1 unit**1 unit Gradient of road PQ 1 = 6 What is the meaning of gradient = 1 : 6? Q P The ratio 1 : 6 means that when the change in vertical distance is 1 unit, the change in horizontal distance is 6 units.**1 unit**1 unit Gradient of road PQ If the inclined road makes an anglewith the horizontal, then 1 = 6 We call the inclination of the road. Q P**vertical**distance = Gradient of a road horizontal distance In general, we have: If an inclined road makes an angle θ with the horizontal, then**Let be the inclination of the ramp.**q tan q 1 = 12 \ ° = ° (cor. to the nearest 0.1 ) q 4 . 8 \ ° The inclination of the ramp is 4.8 . Let’s study the following example. The photo shows a ramp with gradient 1 : 12.**Let be the inclination of the road.**q 1 = Gradient = tan q \ 10 \ ° inclination The of the road is 5.71 . Follow-up question 1 Joe is walking up a steep road of gradient 1 : 10. (a) Find the inclination of the road, correct to the nearest 0.01°. (b) If Joe travels a horizontal distance of 20 m, what is the vertical distance he has travelled? Solution (a)**\**\ The required vertical distance is 2 m. Vertical distance = 2 m Follow-up question 1 (cont’d) Joe is walking up a steep road of gradient 1 : 10. (a) Find the inclination of the road, correct to the nearest 0.01°. (b) If Joe travels a horizontal distance of 20 m, what is the vertical distance he has travelled? Solution (b)**Example 1**Solution**Example 2**Solution**Finding Gradient from a Map**Figure (1) Figure (2) Figure (1) shows the contour map of an island. The curves marked with 100 m, 200 m and 300 m are called contour lines. Any two points along the same contour line (e.g. A and B) have the same vertical distance from sea level (see Figure (2)).**=**- Vertical distance of AC (200 100) m = 100 m = 2 20 000 cm = 40 000 cm = 400 m In the figure, AC is a straight inclined road. If AC is measured to be 2 cm on the map, then horizontal distance of AC Scale = 1 : 20 000 Let’s try to find the gradient of road AC.**=**- Vertical distance of AC (200 100) m = 100 m = 2 20 000 cm = 40 000 cm = 400 m vertical distance = So, gradient of road AC horizontal distance 100 m = 400 m = 1 4 : In the figure, AC is a straight inclined road. If AC is measured to be 2 cm on the map, then horizontal distance of AC Scale = 1 : 20 000**150 m**A 100 m B 50 m scale 1 : 10 000 (a) = - Vertical distance of AB (100 50) m = 50 m = Horizontal distance of AB 2 10 000 cm = 20 000 cm = 200 m Follow-up question 2 A student measures AB to be 2 cm on the contour map as shown in the figure. (a) Find the gradient of road AB in the form 1 : n. (b) Find the inclination of road AB, correct to the nearest 0.01°. Solution Scale = 1 : 10 000**150 m**A 100 m B 50 m scale 1 : 10 000 vertical distance (a) = ∴ Gradient of road AB horizontal distance 50 m = 200 m = 1 : 4 Follow-up question 2 (cont’d) A student measures AB to be 2 cm on the contour map as shown in the figure. (a) Find the gradient of road AB in the form 1 : n. (b) Find the inclination of road AB, correct to the nearest 0.01°. Solution**150 m**A 100 m B 50 m scale 1 : 10 000 Let be the inclination of road AB. q = gradient of AB q tan 1 = 4 \ q = ° ° 14.04 (cor. to the nearest 0.01 ) \ ° The inclination of road AB is 14.04 . Follow-up question 2 (cont’d) A student measures AB to be 2 cm on the contour map as shown in the figure. (a) Find the gradient of road AB in the form 1 : n. (b) Find the inclination of road AB, correct to the nearest 0.01°. Solution (b)**angle of elevation**angle of depression Study the following figure carefully. the angle between our line of sight and the horizontal is called the angle of elevation. When we see an object above us, When we see an object below us, the angle between our line of sight and the horizontal is called the angle of depression.**angle of depression of A from B**angle of elevation of B from A Consider two animals A and B as shown in the figure. Are the angle of elevation of B from A and the angle of depression of A from B equal?**angle of depression of A from B**angle of elevation of B from A Consider two animals A and B as shown in the figure. i.e. are ∠BAD and ∠ABC equal? ∵CB and AD are two horizontal lines. ∴ They are parallel. Also, ∠BAD and ∠ABC form a pair of alternate angles. ∴∠BAD = ∠ABC**angle of depression of A from B**angle of elevation of B from A Refer to the figure. Angle of elevation of B from A = angle of depression of A from B**30** = tan 50 BC 30 = BC tan 50 \ = BC 25.2 cm (cor. to 3 sig. fig.) \ The distance between the flower and the worm is 25.2 cm. In the figure, the height of the flower is 30 cm. The angle of elevation from the worm to the top of the flower is 50. A Consider right-angled triangle ABC. 30 cm C 50 B**AD**Ð = tan ACD DC - AE CF = DC - (60 49) m = 77 m 1 = 7 In the figure, AE and CF are two buildings. Find the(a) angle of elevation of A from C,(b) angle of depression of C from A.(Give your answers correct to 3 significant figures.) Let’s study one more example. A Solution B (a) Consider △ADC. C D 60 m 49 m 77 m E F**1**∵ tan ÐACD = (a) (alt. ∠s, AB // DC) 7 (b) BAC Ð \ The angle of elevation of A from Cis 8.13. =Ð ACD \ = ACD 8.13 (cor. to 3 sig. fig.) Ð \ The angle of depression of C from A is 8.13. = 8.13 In the figure, AE and CF are two buildings. Find the(a) angle of elevation of A from C,(b) angle of depression of C from A.(Give your answers correct to 3 significant figures.) A Solution B C D 60 m 49 m 77 m E F**B**CD Ð = tan CAD AD C 10° CD tan 20° = A 10 m 20° D 10 m Follow-up question 3 The figure shows a monkey and a banana tree. Find the distance between the two bananas at points B and C. (Give you answer correct to 3 significant figures.) Solution Consider △ADC. CD =10 tan 20° m**B**BD Ð = tan BAD AD C 10° BD tan (10°+20°) = A 20° 10 m D 10 m Follow-up question 3 (cont’d) The figure shows a monkey and a banana tree. Find the distance between the two bananas at points B and C. (Give you answer correct to 3 significant figures.) Solution Consider △ADB. BD =10 tan 30° m**B**C 10° A 20° D (cor. to 3 sig. fig.) = 2.13 m 10 m Follow-up question 3 (cont’d) The figure shows a monkey and a banana tree. Find the distance between the two bananas at points B and C. (Give you answer correct to 3 significant figures.) Solution ∴ The distance between the two bananas =BD-CD = (10 tan 30°-10 tan 20°) m**(P.243)**Yes 30o, since 2 angles are alt. angles. They are equal. ID6 (P.243) 39o ID7 (P.244) 170 m**ID8 (P.244)**3800 m ID9 (P.245) ID10 (P.247) h = 50tan60o 10.3 m h =50tan60o -50tan35o) PQ=QB-PB = 80(tan37o-tan32o) = 10.3 m 86.6 m 51.6 m**CP (P.247)**25o 24.6o**CP (P.248)**53 m 12.2 m**The eight directions**Simple Bearings We have learnt how to find directions with a compass in primary school. To describe the directions of a point relative to another point more precisely, true bearing or compass bearing may be used.**P**True Bearing When using the true bearing, directions are measured from the north in a clockwise direction. clockwise direction x° It is expressed as x°, where 0 ≤x <360and the integral part of x must consist of 3 digits.**The true bearing must consist of 3 digits.**For example, A 120° 65° 065° B 273° The true bearings of A, B and C measured from O are 120° 273° 065° , and respectively.**B**A C D Compass Bearing When using the compass bearing, directions are measured from the north or the south. Sx°E, Nx°E, Sx°W or It is expressed as Nx°W, where 0 < x < 90.**For example,**The compass bearing of A from O is N26°E, and the compass bearing of B from O is S65°W.