Transcript of Trigonometrical rules for finding sides and angles in triangles which are not right angled.
- Slide 1
- Trigonometrical rules for finding sides and angles in triangles
which are not right angled
- Slide 2
- AB C First, a word about labelling triangles The vertices
(corners) of a triangle are usually labelled using capital letters,
for example A, B, C The sides of the triangle are usually labelled
using lower case letters, in this case, a, b and c, and are
positioned opposite the respective vertices. So Side a will be
opposite vertex A a Side b will be opposite vertex B b Side c will
be opposite vertex C c IMPORTANT!! Note also that side a could also
be called BC as it connects vertex B to vertex C etc. (We wont be
using this labelling system in this unit of work)
- Slide 3
- We will look at the two rules very briefly before starting to
use them!
- Slide 4
- AB C a b c The Sine Rule states that in any triangle ABC. This
is the general formula for the sine rule. In reality however, you
will use only two of the three fractions at any one time. So the
rule we will be using is More on this later!
- Slide 5
- AB C a b c The Cosine Rule states that in any triangle ABC.
This formula has c 2 as the subject, but the letters can be
interchanged, so it can also be written as or Study the patterns
and locations of the letters in the three formulae closely. More on
the cosine rule later!
- Slide 6
- The Sine Rule
- Slide 7
- Proof of the Sine Rule: Let ABC be any triangle with side
lengths a, b, c respectively C B A c a b Now draw AD perpendicular
to BC, and let the length of AD equal h h In BDCIn ACD and As both
expressions are equal to h, we can say a sin B = b sin A Dividing
through by (sinA)( sinB) this becomes which is the Sine Rule D
- Slide 8
- Example 1 Use the Sine Rule to find the value of x in the
triangle: CA B 88 12m x m 54 VERY IMPORTANT!! Take time to study
the diagram. Note the positions of the three givens (actual values
youre told) the 88 , 54 and 12 m, and the one unknown, x. The
formula for the sine rule requires three givens (in this case, 88 ,
54 and 12 m) and one unknown (x) Note that the third angle C and
its opposite side c are not used in this problem! two of these
givens must be an angle and its opposite side (in this case, the 54
and the 12 m which we will make our A and a). the third given (88)
and the unknown (x) must also be an angle and its opposite
side.
- Slide 9
- CA B 88 12m x 54 Now we substitute the 3 givens and the unknown
into this formula.. A = 54 a = 12 Remember these two givens must be
an angle and its matching opposite side B = 88 b = x Cross-multiply
Divide through by sin 54 to make the subject x = 14.82 (to 2 dec.
pl) Substituting the values into the formula Finally, label the x
as 14.82 on the diagram and check that your answer fits with the
other numbers in the problem! 14.82m (looks OK) These too!
- Slide 10
- Example 2 Use the Sine Rule to find the value of x in the
triangle: 95 35cm x cm 22 Here, no vertices are labelled so we will
have to create our own. But first Step 1, check that there are 4
labels i.e. 3 givens and 1 unknown. There are a 95 , 22 , 35 cm and
x cm so this fits our requirements. Step 2, check that 2 of the 3
givens are a matching angle and opposite side. 95 and 35 cm fit
this. Also check that the remaining given and the unknown form
another matching angle and opposite side (22 and x cm). They do!
All our requirements are in place so we can now use the Sine Rule!
Step 3, Allocate letters A, a, B, b (or any other letters of your
choice) to matching pairs. A = 95 a = 35 B = 22 b = x
- Slide 11
- A = 95 a = 35 B = 22 b = x x = 13.16 (2dec pl) Remember to
check that the answer fits the context of the diagram. 95 35 cm x
cm 22 A a B b
- Slide 12
- Example 3 Use the Sine Rule to find the value of in the
triangle: 62 4.7m 5.1m A quick check indicates everything is in
place to use the Sine Rule. 3 givens and one unknown One pair of
givens (5.1 and 62) form a matching angle and opposite side; and
The other pair (4.7 and ) form the second matching angle and
opposite side. Note the third side and angle are unmarked we dont
use these.
- Slide 13
- 62 4.7m 5.1m Remember to check that the answer fits the context
of the diagram.
- Slide 14
- Example 4 Use the Sine Rule to find the value of x in the
triangle: 68 33 x m 35.7m Looking at the diagram, it seems we have
a problem! Although the 68 and 35.7 form a matching angle and
opposite side, the 33 and x do not. Butremembering the angle sum of
a triangle is 180 , we can work out the 3 rd angle to be 180 33 68
= 79 . So now we use the 79 as the matching angle for the x and
proceed as usual, ignoring the 33 which plays no further part. 79 x
= 37.80 (2 dec pl)
- Slide 15
- Example 5 The Ambiguous Case. Draw two different shaped
triangles ABC in which c = 14m, a = 10m and A = 32 . Hence find the
size(s) of angle C. A B 14m 32 C1C1 10m This process (drawing
triangles from verbal data and no diagram) takes time and practice.
You need to access these types of problems and practise them
thoroughly. Below is one possible diagram: Now extend side AC 1
past C 1 to the new point C 2 where the new length BC 2 is the same
as it was previously (10m).. A B 14m 32 C1C1 10m C2C2 The new ABC 2
has the same given properties as the original ABC 1. Both triangles
have c = 14, a = 10 and A = 32 . But note the angles at C are
different! One is acute and the other obtuse.
- Slide 16
- A B 14m 32 C1C1 10m ANGLE C is obtuse B 14m 32 C1C1 C2C2 10m A
TRIANGLE 1TRIANGLE 2 ANGLE C is acute How are the two C angles
related? (if at all) A B 14m 32 C1C1 10m C2C2 Let angle BC 2 C 1 =
. angle BC 1 C 2 = . (isos ) angle BC 1 A = 180 (straight line) 180
Conclusion: The (green) acute angle at C 2 and the (blue) obtuse
angle at C 1 are supplementary. Thus, for example if one solution
is 73 then the other solution is 180 73 = 107
- Slide 17
- Back to the question! Draw the triangle with the acute, rather
than the obtuse, angle at C. Applying the Sine Rule, B 14m 32 C2C2
10m A One solution (the acute angle which is the only one given by
the calculator) is therefore 47.9 and the second solution (the
obtuse angle) is 180 47.9 = 132.1 Ans: = 47.9 or 132.1
- Slide 18
- The Sine Rule can be used to find unknown sides or angles in
triangles. The Sine Rule formula is To use the Sine Rule, you must
have A matching angle and opposite side pair (two givens) A third
given and an unknown, which also make an angle and opposite side
pair When confronted with a problem where you have to decide
whether to use the Sine Rule or the Cosine Rule, always try for the
Sine Rule first, as it is easier. We will have this discussion
later! When asked to find the size of an ANGLE, first check whether
the problem could involve the ambiguous case (see Example 5). In
that case, the two answers are supplementary i.e. add to 180 In
every triangle, the largest side is always opposite the largest
angle. The side lengths are in the ratio of the sines of their
opposite angles.
- Slide 19
- In every triangle, The largest side is always opposite the
largest angle. The middle sized side is always opposite the middle
sized angle, and The smallest side is always opposite the smallest
angle The ratio of any two side lengths is always equal to the
ratio of the sines of their respective opposite angles. a b c C B A
These are just re-shaped versions of the original sine rule
formulae.
- Slide 20
- The Cosine Rule There are two variations of this. To find a
side use c 2 = a 2 + b 2 2ab cos C To find an angle use These
formulae are just rearrangements of each other. Verify this as an
exercise.
- Slide 21
- Proof of the Cosine Rule: Let ABC be any triangle with side
lengths a, b, c respectively A B C a c b Now draw AD perpendicular
to BC, and let the length of AD equal h h In ACDIn ABD Pythagoras
gives D Let the length CD = x, and so length BD will be a x. xa x
(1) (2) In ACD Pythagoras gives (3) The formulae (2) and (3) are
both for h 2 so we make them equal to each other. NOTE!! The
expansion (a x) 2 = a 2 2ax + x 2
- Slide 22
- Now cancel the x 2 on each side and make c 2 the subject From
the first box on the previous slide, taking result (1) x = b cos C
(4) and substituting this into (4), we get which is a version of
the Cosine Rule (for finding a side)
- Slide 23
- c 2 = a 2 + b 2 2ab cos C (1) Note the positions of the
letters. If the 2ab cos C were missing, this would just be
Pythagoras Theorem, c 2 = a 2 + b 2. If the triangle were right
angled, then C would be 90 and as cos 90 = 0, it becomes Pythagoras
Theorem! (2) When c 2 is the subject, the only angle in the formula
is C (the angle opposite to side c). Note A and B are absent from
the formula. (3) The above formula is to find a side length. The
letters can be swapped around and the same formula can be written b
2 = a 2 + c 2 2ac cos B a 2 = b 2 + c 2 2bc cos A c 2 = a 2 + b 2
2ab cos C Here are the three variations of the formula shown
together. Study them closely and note the patterns!
- Slide 24
- c 2 = a 2 + b 2 2ab cos C (4) This formula can be rearranged to
make cos C the subject, i.e. This is the version of the Cosine Rule
to use when FINDING AN ANGLE. (5) Again, the letters can be swapped
around and the same formula can be written
- Slide 25
- When do we use the Cosine Rule? First, check to see if you can
use the Sine Rule. Its easier! You are told ALL THREE SIDES and
asked to FIND ANY ANGLE You can use the Cosine Rule when OR 8m8m
9m9m 10m You are told TWO SIDES and THEIR INCLUDED ANGLE (i.e. the
angle between those two sides) and asked to FIND THE THIRD SIDE 20
cm 45 15 cm x Here, we use c 2 = a 2 + b 2 2ab cos C
- Slide 26
- Example 6 Use the Cosine Rule to find the value of c in the
triangle: Note that we have 2 given sides (3 cm and 4 cm) and their
included angle (65 ) 65 4 cm c C 3 cm A B c 2 = a 2 + b 2 2ab cos C
Let a = 3 b = 4 C = 65 c 2 = 3 2 + 4 2 2 3 4 cos 65 c 2 = 14.857
(do in one step on calculator) c = 3.85 (to two dec pl) Ans: The
length of the required side is 3.85 cm Finally, check that c = 3.85
fits the diagram. so we can use the Cosine Rule for finding a
side
- Slide 27
- Example 7 Use the Cosine Rule to find the size of C in the
triangle: Note that we have 3 given sides and are asked to find
angle at C (opposite 7.5) 7.5 m 8 m ? B 9 m C A Let a = 8 b = 9 c =
7,5 Ans: Angle C is equal to 51.95 (to 2 dec pl) or 51 57 (to
nearest minute) Finally, check that C = 51.95 fits the diagram. so
we can use the Cosine Rule for finding an angle Caution! Here we
MUST make c = 7.5 as it is the side opposite the angle were
finding, i.e. C, whereas a and b are interchangeable. = 0.6163 NOTE
!! Bracket numerator and denominator when entering into
calculator.
- Slide 28
- Example 8 Use the Cosine Rule to find the value of x in the
triangle: Note that we have 2 given sides (10 m and 11 cm) and
their included angle (100 ) 100 11 m x 10 m c 2 = a 2 + b 2 2ab cos
C Let a = 10 b = 11 c = x C = 100 x 2 = 10 2 + 11 2 2 10 11 cos 100
x 2 = 259.2 (do in one step on calculator) x = 16.10 (to two dec
pl) Ans: The length of the required side is 16.10 m Finally, check
that x = 16.10 fits the diagram. x is the longest side so this
would seem reasonable. so we use the Cosine Rule for finding a
side
- Slide 29
- Example 9 Use the Cosine Rule to find the value of in the
triangle: 29 mm 21 mm 40 mm Note that we have 3 given sides and are
asked to find angle opposite to 40 mm so we use the Cosine Rule for
finding an angle Let a = 21 b = 29 c = 40 C = = 105.13 Ans: is
approx. equal to 105.13 (to 2 dec pl) or 105 8 (to nearest min)
remember the brackets Note the negative cos. This means our angle
is obtuse! ALL OBTUSE ANGLES HAVE A NEGATIVE COSINE! Finally, check
that = 105 fits the diagram. LOOKS obtuse so this would seem
reasonable. Beware you cant always presume the drawings are to
scale, so be careful when judging the appropriateness of your
answers (in all problems)
- Slide 30
- The Cosine Rule can be used to find unknown sides or angles in
triangles. There are two versions of the Cosine Rule formula and
three variations within each of these, depending on what is
required as the subject c 2 = a 2 + b 2 2ab cos C To find a SIDETo
find an ANGLE a 2 = b 2 + c 2 2bc cos A b 2 = a 2 + c 2 2ac cos B
Make sure you familiarise yourself with how the PATTERNS in these
configurations work. Also remember each formula on the left is just
a rearrangement of its corresponding formula on the right.
- Slide 31
- To use the Cosine Rule to find an angle you must be given all
three sides When deciding whether to use the Sine Rule or the
Cosine Rule, always try the Sine Rule first, as it is easier (only
one formula to deal with). To use the Cosine Rule to find a side
you must be given the other two sides and their included angle.
When dealing with angles in the range 90 < < 180 , i.e.
OBTUSE ANGLES, remember that their cosines are negative. This does
not apply to their sines they are still positive.
- Slide 32
- Mixed examples which rule to use? Study each of these diagrams
and determine which rule to use Sine Rule or Cosine Rule? If Cosine
Rule, which version? Answers & working on next slides. A 35 71
x m 16 m E 80 9 cm 6 cm B 14 cm 10 cm 12 cm D 67 x m 11 m 13 m F 33
9 cm x cm 12 cm
- Slide 33
- A 35 71 x m 16 m Example 10 First check to see if we can use
the Sine Rule. We have a given angle and opposite side (35 and
16m), and the unknown x and the other given (71 ) also form a
matching angle and opposite pair. So we can use the SINE RULE to
two dec pl. Ans: the length of side x is 26.38 m approximately.
Remember to check appropriateness of your answer!
- Slide 34
- Example 11 First check to see if we can use the Sine Rule. We
are not given any angle so we cant use the Sine Rule so we have to
use the COSINE RULE the angle version Ans: the size of angle is
44.42 or 44 25 approx. Remember to check appropriateness of your
answer! B 14 cm 10 cm 12 cm Let. C = c = 10 a = 12 b = 14
- Slide 35
- Example 12 First check to see if we can use the Sine Rule. We
have a given angle and opposite side (29 and 12cm), but the unknown
x and the other given (119 ) are NOT a matching angle and opposite
pair. BUTthe third angle is 180 119 29 = 32 so we can use the SINE
RULE to two dec pl. Ans: the length of side x is 13.12 cm
approximately. Remember to check appropriateness of your answer! 32
Let. a = x A = 32 b = 12 B = 29
- Slide 36
- Example 13 First check to see if we can use the Sine Rule. We
are not given any angle and matching opposite side so we cant use
the Sine Rule, so we have to use the COSINE RULE the side version
Ans: the size of side x is 13.35 m (to 2 dec places) Remember to
check appropriateness of your answer! Let. C = 67 c = x a = 11 b =
13 D 67 x m 11 m 13 m c 2 = a 2 + b 2 2ab cos C x 2 = 11 2 + 13 2 2
11 13 cos 67 x 2 = 178.251 x = 13.35
- Slide 37
- Example 14 First check to see if we can use the Sine Rule. We
have a given angle and opposite side (80 and 9 cm), but the unknown
and the other given (6 cm) are NOT a matching angle and opposite
side. HOWEVERwe can use the SINE RULE to find the third angle
(which forms a matching pair with the 6cm) then use the 180 rule to
find Ans: the size of angle is approx. 58.96 or 58 58 Remember to
check appropriateness of your answer! Let. a = 6 A = b = 9 B = 80 E
80 9 cm 6 cm
- Slide 38
- Example 15 F 33 9 cm x cm 12 cm First check to see if we can
use the Sine Rule. We have a given angle and opposite side (33 and
9 cm), but the unknown x and the other given (12 cm) are
insufficient data for Sine Rule. The Cosine Rule wont work either
as the triangles data does not match either of the two
configurations for the Cosine Rule. HOWEVERif we let be the angle
opposite the 12cm we then have a second matching pair and can begin
with using the SINE RULE to find angle . (This is PART 1 ) NOW FOR
PART 2 ..Once we know we can then find the third angle (which is
opposite to x) and then apply the Sine Rule a second time to find
x. Part 1 (finding ) Finding = 180 33 46.57 = 100.43 Part 2
(finding x) Note!! Here the diagram is quite out of scale. This
becomes apparent on checking the reasonableness of your answer