This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ,9 anles & sies of a triangle part 2 This UNIT consists of 2 parts. Part 2: You can calculate the missing angles and sides of any triangle. When you have worked through Part 2, you should;
' know how to relate the Sine Rule to sides and angles of a general triangle * know how to relate the Cosine Rule to sides and angles of a
general triangle  use the Sine Rule and Cosine Rule appropriately to ﬁnd unknown sides
and angles ' be able to interpret the results of calculations intelligently. Prerequisites:
(1) You should be familiar with the material covered in Part 1 of this UNIT (ii) The quadratic equation formula —b :Vbz ~— 4626 2a 1:: to solve the equation ax2 + bx + C = 0 (iii) sin(1800 — x) = sinx
[See UNIT on SIN E, COSINE and TANGENT Part 2] You will need a calculator with the scientic functions SIN},r COS and TAN. Introduction In this Part we calculate sides and angles of a general triangle ABC, which may
contain 3 acute angles (less than 900), or 2 acute angles and 1 obtuse angle (between 900 and 1800). angle B is obtuse We should be aware that the answer may not always be unique. There are two rules in Trigonometry which can be used to solve the problems: I The Sine Rule II The Cosine Rule c2 = a2 + b2 — 2abcosC [other versions of this are a2 = b2 + 6'2  2bccosA and b2 = a2 + c2 — 2accosB Always use the sine rule in preference to the cosine rule, where appropriate,
because of itsrelative simplicity. The problems can be classiﬁed into 3 kinds: Type (i) 2 angles and 1 side are given
Type (ii) 2 sides and 1 angle are given
Type (iii) 3 sides are given. The other sides and/ or angles have to be calculated. Examples Type (i) always use the Sine Rule.
Example 1 If A = 800, B = 300, a =15 find b. b 15 sin 300 sin 800 __ 155m 300
sin 800 b=7.6 b ExampleZ If A = 800, B = 400, .6: =15, find C. First ﬁnd angle C C =1800 —'800 — 400
= 600 Then use the Sine Rule c 15
sin 600 = sin 800
_153in600 _ sin800
6:132. C Type (ii)
To ﬁnd the 3rd side we use the Cosine Rule Example 3 If A = 700 , b = 15, C = 8 ﬁnd a and then calculate angle C The cosine rule gives a2 = 64 + 225 — 240 cos700 the“ a2 = 289 _ 82.08
= 206.92
a=1¢4 Now use the sine rule to ﬁnd C sin C sin 700 8 14.4
' 0
SinC ___ 8311170
14.4
= 0.522
C = 31.50 (Check that angle B is 78.50.) Notice in Example 3 that the given angle is included between the known sides —
this “ﬁxes” the triangle uniquely. In the next example, the angle is not included
between the sides and there is a possibility that the triangle may not be unique. Harder!
Example 4 IfA=659,a=12,c=5findb Using the cosine rule gives 144 = 25 + b2 — 10b 003650 This leads to a quadratic equation in b with 2 possible solutions. We have to
decide whether both are appmpriate. The quadratic equation
b2 — 4.226!) —— 119 = 0
is solved using the formula W
2 b giving solutions b = 15.9 01‘ — 11.7
Clearly, a negative solution would not be allowed for the value of a length. The ambiguity in this example could have been avoided by ﬁrst using the sine
rule to ﬁnd angle C. The angle sum would then give angle A. Finally the cosine
rule could be used to ﬁnd b. Check this by doing the calculations yourself. Type (iii) Use the cosine rule to ﬁnd one of the angles, then the sine rule to
calculate a second angle. Example 5
A ‘ 13 B 10.5 c If 6! =10.5, b = 5, C =13, ﬁnd the angles A, B and C.
Use the cosine rule to calculate angle A (say). 10.52 =13»2 + 52 —13OCosA
ie 110.25 = 169 + 25 — 130 005A ie 130cosA =169+25 —110.25 = 83.75
83.75
cosA = —— = 0.644
hence 130
A = 49. 90 Harder! !
The sine rule then gives sin C _ sin 49.90 13 10.5
' 0
sinC = 135m 49.9
10.5
sinC = 0.947 It would now seem correct to read from the calculator C = 71.20 However. stop and check! It would then follow that B = 580
Can you sense a difﬁculty here? Always check this Put the sides in ascending order
i.e. b=5 a=105 6:13 What order ought the angles to follow?
B , A, C H! What’s gone wrong? If we had used the sine rule to calculate B directly. we should ﬁnd B = 21. 40
[Check this for yourself] The clue to What is going wrong lies in the fact that the sine ratio of an acute
angle and an obtuse angle are both positive. In fact, sin(1800 — 9) = Sin 9
For instance sin 600 = Sin1200 In our example we have The calculation gives just one possible answer C = 71.20 We need to remember that C might be given by
C = 1800 — 71.20
= 108.80 This then gives B = 1800  108.80 — 49. 90
i.e. B = 21.30 and the problem is solved satisfactorily. [Hint We could have avoided the difﬁculty if we had decided to ﬁnd the largest
angle ﬁrst, as the cosine rule gives a unique answer] Exercise In the general triangle ABC
1. IfA= 350, C=1150, C= 21, find a.
2. If B = 400, C = 680, b =13.5, find a.
3. If a = 10.5, C =12, B = 500. find b.
Harder 4. If A = 450, a =14, b = 6.5, find 6. Sketch a diagram to illustrate your results. Harder
5. If a = 25, b = 21, c 2 13 find the angles
(i) calculating the largest angle ﬁrst.
(ii) calculating one of the other angles ﬁrst. _ 215in350
sin1150 Solutions 1. a a 13.3 2. A=720 a=20.0 3. b = 9.6 4. c2 — 25.37% + 123.75 = 0 The cosine rule leads to the equation Solving, C = 6.6 01‘ 18.8 Two possible triangles may be drawn. Draw them to check. 5. A = 91.60(obtuse angle), B = 57.10, C = 31.30. ...
View
Full
Document
This note was uploaded on 03/16/2011 for the course ECON 1003 taught by Professor Dr.mcdonald during the Spring '11 term at University of the West Indies at Mona.
 Spring '11
 DR.MCDONALD

Click to edit the document details