1. h) select and use appropriate trigonometric ratios and formulae to solve problems involving trigonometry that require the use of more than one triangle (two dimensions), where the diagram is provided. Non-right-angled triangles and trigonometry e) solve angle and length problems using the sine rule in acute- and obtuse-angled triangles d) draw graphs of the sine and cosine curves for 0°  A  180° f) solve angle and length problems using the cosine rule in acute- and obtuse angled triangles Cosine rule: c 2 = a 2 + b 2 – 2ab cos C, and g) use the area formula to find the area of a triangle Area =½ab sin C (All these things are on your self-assessment sheet for this topic) a) find the trigonometric ratios of obtuse angles either from redefinition of the trigonometric relationships using the circle (usually the unit circle) or by use of a calculator b) find the possible acute and/or obtuse angles, given a trigonometric ratio c) establish and use the following relationships for obtuse angles (0°  A  90°): sin (180° – A) = sin A cos (180° – A) = – cos A tan (180° – A) = – tan A 

2. Finding the length of sides and size of angles in non right-angled triangles So far in your trig work, you have only used acute angles, because the triangles you have been working with have been only right-angled. What do you do in these triangles? Why can’t you use normal Sin, Cos or Tan Trigonometry? There is no Hypotenuse because they aren’t right-angled triangles! So how do you find the side marked x or the angle marked  ? 12 m x m 35° 110° Find x. 15 m 28 m  25° Find . A new rule is needed! 

3. Let’s investigate these types of triangles! A B C First of all, we need to be able to name the sides of the triangle. We could use two letters, e.g. AB, but it would be quicker if it was only one letter! The convention used is that the side opposite an angle(which is ALWAYS named with an upper case letter) is named by a lower case of the same letter. So, this side is a This side is b And, this side is c

4. A B C ab c In this  ABC, to develop the new rule, first draw a perpendicular from C to AB. Name the intersecting point D and the line h. Do it now on your diagram In  ACD Sin A = h = b  Sin A b D h And, by manipulating, h Now,  ABC is broken up into 2 triangles,  BCD &  ACD, (and they are both right angled). And in  BCD Sin B = h = a  Sin B a And, by manipulating, h We now have 2 statements which are both equal to h. So, we can make them equal to each other…. b Sin A And, by manipulating, b Sin BSin A a = a Sin B=

5. So, we have derived a rule which uses 2 sides and their opposite angles. b Sin BSin A a = A B C ab c D h A B C a b c D h If we changed the perpendicular so that it came from A and went to CB, we would come up with this rule: c Sin CSin B b = Both parts of the two rules can be combined to develop what is known as the Sine rule: c Sin CSin B b = Sin A a =

6. When the rule is in this form, we shall first use it to find a side: First,because the vertices of the triangle have not been named, name each corner. It doesn’t matter where you put each of the vertex letters, but it DOES matter where you put the side letters! c Sin CSin B b = Sin A a = But we will only ever need 4 of the six parts of the rule in any one use of the formula. A B C a b c As we don’t know a, we won’t be able to use it OR angle A. Having placed the letters, substitute into the relevant parts of the formula correctly. c Sin CSin B b = Sin A a = x Sin 40° = 9 Sin 80° x =x = 9  Sin 40° Sin 80° And, by manipulating, x x  5.87 cm (to 2 d.p.) This is a reasonable answer as it opposite the 40°, and the 9cm is opposite the 80° Don’t presume that just because 40° is half 80° that the answer must be half 9cm! ( ) 80° 40° x cm 9 cm Find x to 2 decimal places.

7. c Sin CSin B b = Sin A a = Once again we will only ever need 4 of the six parts of the rule in any one use of the formula. a b c This time, we don’t know b, so we won’t be able to use it OR angle B. Having placed the letters, substitute into the relevant parts of the formula correctly. = Sin  ° 18 Sin 10° 6 Sin   18  Sin 10° 6 And, by manipulating,    31  24 Using the INV key on your calculator This is a reasonable answer as it opposite the 19cm, and the 6cm is opposite the 10° Don’t presume that just because 6cm is one third 18cm that the answer must be 3 times 10°! ( ) It will make our work much easier to find an angle if we turn the Sine rule upside down, so that the angles are on the top. After all when we found the sides they were on top! c Sin CSin B b = Sin A a = AB C It doesn’t matter which pair of the information is written first, as long as the information in each fraction is opposite each other in the triangle. Here C and c have been written first. Sin   0.52094…. 10° °° 6 cm 18 cm Find  to the nearest minute. Your turn, Exercise 9F, page 344 Year 10 Text