Presentation is loading. Please wait.

Presentation is loading. Please wait.

Further Trigonometry Sin Opposite Cos Tan Adjacent Hypotenuse Opposite Adjacent Sin Opposite Cos Tan Adjacent Hypotenuse Opposite Adjacent The Cosine.

Similar presentations


Presentation on theme: "Further Trigonometry Sin Opposite Cos Tan Adjacent Hypotenuse Opposite Adjacent Sin Opposite Cos Tan Adjacent Hypotenuse Opposite Adjacent The Cosine."— Presentation transcript:

1

2 Further Trigonometry Sin Opposite Cos Tan Adjacent Hypotenuse Opposite Adjacent Sin Opposite Cos Tan Adjacent Hypotenuse Opposite Adjacent The Cosine Rule

3 We can derive the COSINE RULE as follows: A C B a c b Consider the triangle ABC below: Name the 3 sides of  ABC P Draw CP perpendicular to ABLet AP = x so PB = c - xNow use Pythagoras in  ACPUse Pythagoras in  BCP CP 2 = b 2 – x 2 CP 2 = a 2 – (c-x) 2 Now combine the two parts CP 2 = a 2 – (c-x) 2 b 2 – x 2 Square out the brackets  b 2 – x 2 = a 2 – (c 2 –2cx +x 2 ) Simplify the whole thingWrite x in terms of A, B, or C!! xc - x  b 2 – x 2 = a 2 – c 2 + 2cx - x 2  b 2 = a 2 – c 2 + 2cx Cos A = Adj Hyp Cos A = xbxb x = b x Cos A  b 2 = a 2 – c 2 + 2cbCosA  b 2 = a 2 – c 2 + 2bcCosA Rearrange so that a 2 = ………  b 2 + c 2 = a 2 + 2bcCosA  b 2 + c 2 - 2bcCosA = a 2  a 2 = b 2 + c 2 - 2bcCosA This is the COSINE RULE which can be used to find missing sides in NON RIGHT ANGLED triangles!!

4 Finding a missing side 1: Look at the triangle below: a 2 = b 2 + c 2 - 2bcCosA 4 8m 60 o x Label the vertices and sides A C B a b c Write the FULL Cosine rule a 2 = b 2 + c 2 - 2bcCosA Substitute in all the values Calculate as much as you can x = 6.928m x 2 = 8 2 + 4 2 – 2(8)(4)Cos60 x 2 = 64 + 16 – 64 x 0.5 x 2 = 48 x = 48 

5 Finding a missing side 2: Look at the triangle below: a 2 = b 2 + c 2 - 2bcCosA Label the vertices and sides A B C a c b Write the FULL Cosine rule a 2 = b 2 + c 2 - 2bcCosA Substitute in all the values Calculate as much as you can m = 13.565cm m 2 = 12.7 2 + 6 2 – 2(12.7)(6)Cos85 m 2 = 161.29+36 – 152.4 x 0.0872 m 2 = 184.007 m = 184.007  m 6cm 85 o 12.7cm

6 Finding a missing side 3: Look at the triangle below: HELP!!!! We must find angle A to be able to calculate the missing side using the Cosine rule. A = 180 – 15 – 145 = 165 – 145 = 20 o B C A b a c a 2 = b 2 + c 2 - 2bcCosA p = 25.01m p 2 = 45 2 + 62 2 – 2(45)(62)Cos20 p 2 = 2025+3844 – 5580 x 0.9397 p 2 = 625.515 p = 625.515  15 o p 145 o 45 m 62 m 20 o

7 The Cosine Rule The cosine rule in this form: a 2 – (b 2 + c 2 ) = - 2bcCosA Take b 2 + c 2 from both sides  b 2 + c 2 - a 2 = 2bcCosA Multiply both sides by -1Divide both sides by 2bc  b 2 + c 2 - a 2 = CosA 2bc Rearrange so that CosA = ……… a 2 = b 2 + c 2 - 2bcCosA This is the COSINE RULE which can be used to find missing angles in NON RIGHT ANGLED triangles!! cannot be used easily for finding a missing angle. We must therefore rearrange it so that CosA = ….. a 2 = b 2 + c 2 - 2bcCosA  b 2 + c 2 - a 2 2bc CosA =

8 Cosx = 0.714….. x = Cos -1 0.714….. Finding a missing angle 1: Look at the triangle below: 4 x o 7m 5m Label the vertices and sides A C B a b c Write the FULL Cosine rule Substitute in the all values Calculate as much as you can b 2 + c 2 - a 2 2bc CosA = b 2 + c 2 - a 2 2bc CosA = 7 2 + 4 2 – 5 2 2(7)(4) Cosx = 49 + 16 – 25 56 Cosx = 40 56 Cosx = x = 44.415 o

9 Cosx = 0.75 x = Cos -1 0.75 Finding a missing angle 2: Look at the triangle below: Label the vertices and sides A C B a b c Write the FULL Cosine rule Substitute in the all values Calculate as much as you can b 2 + c 2 - a 2 2bc CosA = b 2 + c 2 - a 2 2bc CosA = 10 2 + 12 2 – 8 2 2(10)(12) Cosx = 100 + 144 – 64 240 Cosx = 180 240 Cosx = x = 41.410 o x o 10cm 8.cm 12cm

10 53º N Cosine Rule Problem 1: A ship’s distress flare is spotted by a Cruise ship and a lighthouse keeper. Distance between CS and L is 4km Distance from Lighthouse is 9km Bearing of CS from L’house is 053º Bearing of Ship from L’hse is 091º 9km Calculate the distance the Cruise ship has to go to reach the ship? L C S x Angle L = 91 o - 53 o = 38 o 38º l c s 4km l 2 = c 2 + s 2 – 2csCosL x = 6.35km x 2 = 9 2 + 4 2 – 2(9)(4)Cos38 x 2 = 81 + 16 – 72 x 0.788… x 2 = 40.263…… x = 40.263… 

11 Cosine Rule Problem 2: An Army gunner on a hilltop spots an enemy chopper advancing on his camp. It is 2.3km from the gunner. The gunner is 1.7km from camp. The chopper stops 1.2km from camp. x 2.3 km h g C H G c y 1.7 km 1.2 km If the camp is 52º below the gunner what is the angle of depression needed for the gunner to destroy the chopper. h 2 + c 2 - g 2 2hc CosG = 1.7 2 + 2.3 2 – 1.2 2 2(1.7)(2.3) Cosy = 2.89 + 5.29 – 1.44 7.82 Cosy = Angle x = 52 o – 30.470 o = 21.53 o x Cosy = 0.8619 y = Cos -1 0.8619 6.74 7.82 Cosy = y = 30.470 o

12 Which to use? We have now seen how to use the: Cosine Rule: a 2 = b 2 + c 2 - 2bcCosA b 2 + c 2 - a 2 2bc CosA = Sine Rule: a b. c. Sin A Sin B Sin C == Sin A Sin B Sin C a b c == But how do you know which one to use? Thee are 2 ways:

13 Which to use? 1 Memorise the following: Finding a Side:   COSINE RULE: a 2 = b 2 + c 2 - 2bcCosA x  x    SINE RULE: a b. c. Sin A Sin B Sin C == x Finding an Angle:    b 2 + c 2 - a 2 2bc CosA = COSINE RULE: x    SINE RULE: Sin A Sin B Sin C a b c == Which to use? 2 OR always start with the SINE Rule: a b. c. Sin A Sin B Sin C ==     If you can eliminate one part of the SINE rule then use it!!! a b. c. Sin A Sin B Sin C ==    If you cannot eliminate one part of the SINE rule then use the COSINE rule!!!


Download ppt "Further Trigonometry Sin Opposite Cos Tan Adjacent Hypotenuse Opposite Adjacent Sin Opposite Cos Tan Adjacent Hypotenuse Opposite Adjacent The Cosine."

Similar presentations


Ads by Google