Sine and Cosine rules: Sine and Cosine rules 1 Trigonometry applied to triangles without right angles.
Introduction: Introduction You have learnt to apply trigonometry to right angled triangles. 2 A hyp adj opp
PowerPoint Presentation: Now we extend our trigonometry so that we can deal with triangles which are not right angled. 3
PowerPoint Presentation: First we introduce the following notation. We use capital letters for the angles, and lower case letters for the sides. 4 Q q p r R P A a b c C B In D ABC The side opposite angle A is called a . The side opposite angle B is called b . In D PQR The side opposite angle P is called p . And so on
PowerPoint Presentation: There are two new rules. 5
1. The sine rule: 1. The sine rule Draw the perpendicular from C to meet AB at P. 6 A a b C c Using D BPC: PC = a sinB. B P Using D APC: PC = b sinA. Therefore a sinB = b sinA. Dividing by sinA and sinB gives: In the same way: Putting both results together: The proof needs some changes to deal with obtuse angles.
Example 1: Example 1 Find the length of BC. 7 Substitute A = 35 o , B = 95 o , b = 6.2: Multiply by sin35 o : A a 6.2 cm c C B 35 o 95 o
Example 2: Example 2 Find the size of angle R. 8 Q q 18 cm 15 cm R P 100 o Substitute: P = 100 o , p = 18, r = 15: Multiply by 15: leading to leading to (or , impossible here).
PowerPoint Presentation: Try to avoid finding the largest angle as you probably do not know whether you want the acute or the obtuse angle. The largest angle is opposite the longest side, the smallest angle opposite the shortest side. When there are two possible angles they add up to 180 degrees. 9 When using the sine rule to find an angle:
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2. The cosine rule: 2. The cosine rule 11 A a b c C B
Proof of the cosine rule: Proof of the cosine rule Applying Pythagoras’ Theorem to D APC gives: h 2 = b 2 – x 2 j Applying Pythagoras’ Theorem to D BPC gives: a 2 = h 2 + ( c – x ) 2 = h 2 + c 2 – 2 cx + x 2 . k Substituting from equation j into equation k gives: a 2 = b 2 – x 2 + c 2 – 2 cx + x 2 = b 2 + c 2 – 2 cx . l Using D APC again: x = b cosA . Substituting this into l gives: a 2 = b 2 + c 2 – 2 cb cosA . i.e. 12 c A a b C B a 2 = b 2 + c 2 – 2 bc cosA P h c – x x Again the proof needs some changes to deal with obtuse angles. Press to skip proof
There are two main ways of writing the cosine rule: There are two main ways of writing the cosine rule one for finding a side, one for finding an angle. 13
The cosine rule for finding a side.: The cosine rule for finding a side. The formula starts and ends with the same letter, one lower case, one capital. The square of a side is the sum of the squares of the other 2 sides minus twice the product of the 2 known sides and the cosine of the angle between them. 14 A a b c C B
The cosine rule for finding an angle.: The cosine rule for finding an angle. The cosine of an angle of a triangle is the sum of the squares of the sides forming the angle minus the square of the side opposite the angle all divided by twice the product of first two sides. 15 A a b c C B Starting from: Add 2 bc cosA and subtract a 2 getting Divide both sides by 2 bc : D d r m M R
Example 3: Example 3 Find the length of BC. 16 A a 8.3 cm 7.2 cm C B 38 o Substitute A = 38 o , b = 8.3, c = 7.2 into (You need to show what you are calculating, but you do not need to show intermediate results.)
Example 4: Example 4 Find the size of angle D. 17 R 4 cm 8 cm 6 cm M D Substitute d = 8, r = 4, m = 6 The cosine rule automatically takes care of obtuse angles. into getting leading to There is no need to show intermediate results.
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How do I know whether to use the sine rule or the cosine rule?: How do I know whether to use the sine rule or the cosine rule? To use the sine rule you need to know an angle and the side opposite it. You can use it to find a side (opposite a second known angle) or an angle (opposite a second known side). To use the cosine rule you need to know either two sides and the included angle or all three sides. 19
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The Area Of A Triangle.: The Area Of A Triangle. A o B o C o a b c
Finding The Formula.: Finding The Formula. Consider the triangle below: A o B o C o a b c By drawing an altitude h we can calculate the area of the triangle. h Area = ½ x base x height What does the sine of A o equal ? Change the subject to h. h = bsinA o Substitute into the area formula.
Using The Formula.: Using The Formula. Example1 Find the area of the triangle below: 39 o 12m 7m Write down the area formula: Identify b,c and A o b = 7 c = 12 A o = 39 o Substitute the values into the formula. Calculate the area. A = 42 x 0.629 A = 26.4cm 2
PowerPoint Presentation: Example2 Find the area of the triangle below: 123 o 19m 13m Write down the area formula: Identify b,c and A o b = 13 c = 19 A o = 123 o Substitute the values into the formula. Calculate the area. A = 123.5 x 0.839 A = 103.6 cm 2
What Goes In The Box ?: What Goes In The Box ? Calculate the areas of the triangles below: (1) 23 o 15cm 12.6cm (2) 71 o 5.7m 6.2m (3) 151 o 17mm 9mm A =36.9cm 2 A =16.7m 2 A =37.1mm 2
PowerPoint Presentation: Reversing the Process…. Sometimes questions give us the area but ask us to find either angles or sides. Example 2 : In triangle ABC, a = 3.6 cm, b = 4.9 cm and the area is 5 cm². Find the size of angle BCA. Hint: Always sketch a diagram whether you have been given one or not.
PowerPoint Presentation: Example 3 : In triangle ABC, b = 13 cm, Angle CAB = 66° and the area is 100 cm². Find the length of side ‘c’. A B C 13 cm 66° Area = 100cm²
PowerPoint Presentation: Exam Style Question Use the measurements given to find the area of this quadrilateral shaped field. Give your answer to 3sf. 70º 62º 45m 50m 60m 25m
PowerPoint Presentation: 29 Worksheet Time