ADDMATH FORM 4
April 25, 2017 | Author: Nor Hanina | Category: N/A
Short Description
SOLUTION OF TRIANGLES...
Description
SOLUTION OF TRIANGLES 3.2 Solve problems involving three-dimensional objects Task (1)
: Answer all the questions below. H
G
G
Solution: HC= 6 2 32 6.708
E
F
BD = 6 2 4 2 7.211 D
HB = 4.692 32 7.810
C B
A
cos BHC The diagram above shows a cuboid with a rectangular base, ABCD. Given that AB = 6cm, BC = 4cm and CG=3 cm. Find BHC (2) H
= 0.8589 BHC = 30.810
G
E
F
D
A
7.810 2 6.708 2 4 2 2(7.810 )( 6.708 )
C
B The diagram above shows a cuboid with a rectangular base ABCD. Given that AB = 16 cm, BC = 4cm and CG=13 cm. Find BHC [10.98]
(3)
H
E
G
F D
C
B
A
The diagram above shows a cuboid with a rectangular base ABCD. Given that AB = 6 cm, BC = 4cm and CG = 3 cm. Find BGD. [74.44]
Solutions of Triangles
151
4.
H
G
F
D
E
The diagram on the left shows a cuboid with a rectangular base ABCD. Given that DG=6.1 cm, BG=7.2 cm and BGD =41.02.
C
A
Find the length of BD.
B
[ 4.772 cm ]
5.
H
E
G
D
C
F
A
The diagram on the left shows a cuboid with a rectangular base ABCD. Given that BC = 8.2 cm, CG = 6.42 cm, AB = 12.03 cm and ABG =110.02. Find the length of AG.
B
[ 17.91 cm ]
Solutions of Triangles
152
To find unknown sides :
To find unknown angles :
a b c sin A sin B sin C
sin A sin B sin C a b c
10. SOLUTION OF TRIANGLES 1.2 Use Sine Rule to find the unknown sides or angles of a triangle. Task 1 : Find the unknown sides of a triangle when two of its angles and one of the corresponding sides are known. (1) Diagram 1 shows the triangle ABC. Answer :
BC 8.2 0 sin 75 sin 400 BC Diagram 1
8.2 sin 750 sin 400
Using the scientific calculator, BC = 12.32 cm
Calculate the length of BC. (2) Diagram 2 shows the triangle PQR
Diagram 2 [ 8.794 cm ]
Calculate the length of PQ. (3) Diagram 3 shows the triangle DEF. D 15 cm 600 E
350 16’ Diagram 3
F
Calculate the length of DE.
[ 10.00 cm ]
(4) Diagram 4 shows the triangle KLM. L K 420 0 63 15 cm Diagram 4 M Calculate the length of KM. Solutions of Triangles
153
[ 11.26 cm ]
(5) Diagram 5 shows the triangle ABC.
Answer :
ABC 1800 400 750 650
AC 8.2 0 sin 65 sin 400 BC Diagram 5
8.2 sin 650 0 sin 40
Using the scientific calculator, Calculate the length of AC. AC = 11.56 cm (6) Diagram 6 shows the triangle PQR
Diagram 6
Calculate the length of PR. [ 6.527 cm ]
(7) Diagram 7 shows the triangle DEF. D 15 cm 600 E
350 16’ Diagram 7
F
Calculate the length of EF.
[ 17.25 cm ]
(8) Diagram 8 shows the triangle KLM. L K 420 0 63 15 cm Diagram 8 M Calculate the length of KL.
Solutions of Triangles
154
[ 16.26 cm ]
Task 2 : Find the unknown sides of a triangle when two of its angles and the side not corresponding to the angles are known. (9) Diagram 9 shows the triangle ABC. Answer :
ABC 1800 470 780 550
BC 11.2 0 sin 47 sin 550 BC
11.2 sin 47 0 sin 550
Diagram 9 Using scientific calculator, Calculate the length of BC. BC = 9.9996 cm or 10.00 cm (10) Diagram 10 shows the triangle ABC.
Diagram 10 Calculate the length of AC. [ 4.517 cm ]
(11) Diagram 11 shows the triangle PQR. 7.2 cm P
250
0
28
R
Diagram 11 Q Calculate the length of PQ. [ 3.810 cm ]
(12) Diagram 12 shows the triangle DEF. D
720 E
510 5.6 cm
F
Diagram 12 Calculate the length of DE. Solutions of Triangles
155
[ 5.189 cm ]
Task 3 : Find the unknown angles of a triangle when two sides and a non-included angle are given. (1) Diagram 1 shows the triangle ABC. Answer : A 10 cm
sin C sin 60 0 10 15
15 cm
10 sin 60 0 sin C 15
600 B
C
Diagram 1
sin C 0.5774 C sin 1 0.5774 C 35.270
Find ACB.
(2) Diagram 2 shows the triangle KLM 15 cm
L
K 500
9 cm
Diagram 2 M Find KLM [ 27.360 ]
(3) Diagram 3 shows the triangle DEF. D 3.5 cm
12.5 cm
430 24’ E
F
Diagram 3
Find DFE. [ 11.090 ]
(4) Diagram 4 shows the triangle PQR. 13 cm
R
P 10 cm 0
130
Diagram 4 Q Find QPR.
Solutions of Triangles
156
[ 36.110 ]
(5) Diagram 5 shows the triangle ABC.
Answer :
sin A sin 110 0 9 14 9 sin 110 0 sin A 14
A 14 cm
1100 Diagram 5
B
sin A 0.6041 A sin 1 0.6041 A 37.160
9 cm
C
Find ABC.
ABC 180 0 110 0 37.16 0 32.84 0 (6) Diagram 6 shows the triangle KLM. 4.2 cm
K
L
2.8 cm
250
Diagram 6
M Find KLM. [ 138.640 ]
(7) Diagram 7 shows the triangle DEF. E
340
D
4.4 cm
6.7 cm F
Diagram 7
Find DFE.
[ 124.460 ]
(8) Diagram 8 shows the triangle PQR. P
12.3 cm R
550 7.7 cm
Q
Diagram 8
Find PQR.
Solutions of Triangles
157
[ 94.150 ]
Task 4 : Find the unknown side of a triangle when two sides and a non-included angle are given. (1) Diagram 1 shows the triangle ABC. Answer :
sin C sin 37 0 14 9 14 sin 37 0 sin C 9
A 370
14 cm
B Diagram 1
C
sin C 0.9362
9 cm
C sin 1 0.9362
Given that ACB is an obtuse angle, find the length of AC.
C 110.580 B 1800 110.580 370
32.420 AC sin 32.42 0
9
sin 37 0 9 sin 32.42 0 AC sin 37 0 AC = 8.018 cm
(2) Diagram 2 shows the triangle KLM 9 cm
L
K
400 7 cm M
Diagram 2
Given that KLM is an obtuse angle, find the length of ML.
[ 2.952 cm ]
Solutions of Triangles
158
(3) Diagram 3 shows the triangle DEF. D 8 cm 420 E
11 cm
F Diagram 3
Given that the value of EDF is greater than 900, find the length of DE.
[ 5.040 cm ]
(4) Diagram 4 shows the triangle PQR. 8.5 cm P
460
R
6.9 cm Diagram 4 Q Given that PQR is an angle in the second quadrant of the cartesian plane, find the length of QR.
[ 2.707 cm ]
(5) Diagram 5 shows the triangle KLM. K
L 230
17.3 cm
9.2 cm
Diagram 5 M Given that KLM is an angle in the second quadrant of the cartesian plane, find the length of KL.
[ 9.686 cm ]
Solutions of Triangles
159
10. SOLUTION OF TRIANGLES 2.2 Use Cosine Rule to find the unknown sides or angles of a triangle. 2
2
b2 c2 a2 cos A 2bc 2 a c2 b2 cos B 2ac 2 a b2 c2 cos C 2ab
2
a = b + c – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C
Task 1 : Find the unknown side of a triangle when two sides and an included angle are given. (1) Diagram 1 shows the triangle PQR such Solution : that PR =12.3 cm , QR =16.4 cm and PRQ = 67 . x 2 16 .4 2 12 .3 2 2(16 .4)(12 .3) cos 67 0 P = 262.1
12.3 cm
x cm
x 262 .61
670
Q
R
16.4 cm
x 16.21
Diagram 1
Find the value of x. (2) Diagram 2 shows the triangle PQR such that PQ =7 cm, QR =5 cm and PQR = 75 . R
x cm
5 cm 5cm 750 Q
P
7 cm
Diagram 2
Find the value of x. [ 7.475 ]
(3) Diagram 3 shows a triangle with sides 5 cm , 13 cm and an included angle 43 .
x cm
5 cm 430 E
13 cm Diagram 3
Find the value of x .
[ 9.946 ]
Solutions of Triangles
160
(4) Diagram 4 shows the triangle PQR. 7 cm C A 0 53 6.3 cm Diagram 4 B Find the length of BC.
[ 5.967 cm ]
(5) Diagram 5 shows the triangle KLM. 5.8 cm K
L
480
4 cm Diagram 5
M
Find the length of LM. [ 4.312 cm ]
(6) Diagram 6 shows the triangle PQR. R 2.23 cm 750 31’ P
Q 5.40 cm Diagram 6
Find the length of PR. [ 5.302 cm ]
(7) Diagram 7 shows a triangle with sides 6.21 cm , 10.51 cm and an included angle 360 39’ . x cm
6.21cm 360 39′ 10.51cm Diagram 7
Find the value of x .
[ 6.656 ]
Solutions of Triangles
161
Task 2 : Find the unknown angle of a triangle when three sides are given. (1) In Diagram 1, ABC is a triangle where Solution : AB = 13 cm, AC = 14 cm and BC= 15 cm. 13 2 14 2 15 2 cos BAC A 2(13)(14) 14 cm
13cm
B
=0.3846
BAC 67.38
C
15 cm
Diagram 1 Find BAC . (2) Diagram 2 shows a triangle ABC where AB = 11 cm, AC = 13 cm and BC= 16 cm. A 13 cm
11cm
B
C
16 cm
Diagram 2 Find BAC . [ 83.17]
(3) Diagram 3 shows a triangle ABC where AB = 13 cm, AC = 16 cm and BC = 17.5 cm. A 16 cm
13cm
B
17.5 cm
C
Diagram 3 Calculate BAC
[ 73.41]
(4) Diagram 4 shows a triangle ABC where AB = 12.67 cm, AC = 16.78 cm and BC= 19.97 cm. A
16.78 cm
12.67cm
B
19.97 cm
C
Diagram 4 Calculate BCA [39.17]
Solutions of Triangles
162
(5) In Diagram 5, PQR is a triangle such that PR = 6.45 cm, RQ = 2.23 cm and PQ = 5.40 cm. R
6.45 cm
2.23 cm Q
P
5.40 cm Diagram 5 Find RQP . [108.07]
(6) In Diagram 6, PQR is a triangle such that PR = 23.5 cm, RQ = 12.5 cm and PQ= 18.7 cm. R
23.5 cm
12.5 cm Q
P
18.7 cm Diagram 6
Calculate the smallest angle in the triangle. [31.96]
(7) For triangle ABC in Diagram 7, AB = 8.56 cm, AC = 11.23 cm and BC= 14.51 cm. A 11.23cm
8.56cm
B
C
14.5 1cm
Diagram 7 Calculate the largest angle in the triangle. [93.33]
(8) For triangle ABC in Diagram 8, AB = 13 cm, AC = 16 cm and BC= 17.5 cm. A 16 cm
13cm
B
17.5 cm
C
Diagram 8 Calculate the second largest angle in the triangle. [61.19]
Solutions of Triangles
163
Area of ∆ =
10. SOLUTION OF TRIANGLES
1 bc sin A 2 1 = ac sin B 2
1 Use the formula ab sin C or its equivalent to find the area 2
3.1
1 ab sin C 2
=
of a triangle. Task : Find the area of a triangles given in each of the following.. (1) In Diagram 1, ABC is a triangle with Solution: AB= 6 cm, AC = 9 cm and BAC 53 . Area of ABC
A 53
6 cm
9 cm
1 (6)(9) sin 53 2
= 21.56 cm2
B
Diagram 1 C
Find the area of ABC (2) In Diagram 2, ABC is a triangle with AC= 6 cm, BC = 5 cm and ACB 78 . A
6 cm 78
B 5 cm
C
Find the area of ABC.
Diagram 2 2
[ 14.67 cm ]
(3) In Diagram 3, ABC is a triangle with AC= 6 cm, BC = 8 cm and ACB 120 . B 8 cm 120
C 6 cm
A
Diagram 3 Find the area of ABC. (4) In Diagram 4, ABC is a triangle with AC= 6 cm, BC = 12.5 cm and the reflex angle ACB 250 .
2
[ 20.78 cm ]
B
12.5 cm 250
C
6 cm
Diagram 4 A 2
Find the area of ABC. Solutions of Triangles
[ 35.24 cm ]
164
(5) In Diagram 5, ABC is a triangle such that AB= 12.5 cm , AC = 6 cm and ACB=80.
Solution: (a)
C
B
12.5 6 sin CBA sin 80
80 6 cm
sin CBA =
12.5 cm
6 sin 80 12.5
= 0.4727
A
CBA =sin -1 (0.2727)
Diagram 5
=28.21
Find (a) CBA, (b) the area of the triangle.
(b)
CAB 180 28.21 80 =71.79
Area of ABC=
1 (6)(12.5) sin 71.79 2
=35.62 cm
2
(6) In Diagram 6, ABC is a triangle such that AB= 11 cm , AC = 15 cm and ACB=4534’.
A 11cm 15 cm B 4534' C
Diagram 6
Find (a) CBA, (b) the area of the triangle.
[ (a) 76.830 (b) 69.66 cm2 ]
(7) In Diagram 7, ABC is a triangle such that AC = 7 cm, AB = 15 cm and ACB = 11530’. A 15 cm 7cm
11530' C
B
Diagram 7 Find (a) CBA, (b) the area of the triangle
[ (a) 24.910 (b) 33.46 cm2 ]
Solutions of Triangles
165
(8) In Diagram 8, ABC is a triangle where AB= 15 cm, BC =11 cm and AC=8 cm. 11 cm
C
Solution (a)
B
cos B
112 152 8 2 2(11)(15)
=0.8545
8cm
B = 31 30’
15 cm
A
Diagram 8
(b)
Area of ABC =
1 (11)(15) sin 31 30' 2
= 42.86
Find (a) the smallest angle, (b) the area of ABC. (9) In Diagram 9, ABC is a triangle where AB= 30 cm, BC =25 cm and AC=20 cm.
C 25 cm
20 cm
B 30 cm
A
Diagram 9 Find (a) the largest angle, (b) the area of ABC.
0
2
[ (a) 82.82 (b) 248.04 cm ]
(10) In Diagram 10, ABC is a triangle where AB = 13 cm, AC = 14 cm and BC= 15 cm. A 14 cm
13cm
B
15 cm
C
Diagram 10 Find (a) the second largest angle, (b) the area of ABC.
0
2
[ (a) 59.49 (b) 84.00 cm ]
Solutions of Triangles
166
10. SOLUTION OF TRIANGLES Further Practice with questions based on SPM format. Task : Answer all the questions below. (1) Diagram 1 shows a trapezium LMNO. L
13 cm
M
16 cm 31o O
18 cm
N
Diagram 1 Calculate (a) LNM, (b) the length of LN, (c) the area of ∆OLN. 0
2
[ (a) 24.74 (b) 25.67 cm (c) 118.99 cm ]
(2) In Diagram 2, BCD is a straight line. A 32o
10 cm
7 cm
B
C
5 cm
D
Diagram 2 Find (a) ACD, (b) the length of BC, (c) the area of triangle ABD. 0
2
0
2
[ (a) 111.80 (b) 3.769 cm (c) 28.50 cm ]
(3) In Diagram 3, FGH is a straight line and G is the midpoint of FH. E
14 cm
F
16 cm
10 cm
G
H
Diagram 3 Find (a) EFG, (b) the length of EG, (c) the area of triangle EGH.
[ (a) 52.62 (b) 11.23 cm (c) 52.62 cm ]
Solutions of Triangles
167
(4) Diagram 4 shows a quadrilateral KLNM.
Diagram 4 Calculate (a) the length of LM, (b) MNL, (c) the area of quadrilateral KLNM.
0
2
[ (a) 12.92 cm (b) 31.73 (c) 141.65 cm ]
(5) In Diagram 5, QRS is a straight line.
Diagram 5 Find (a) QPR, (b) the length of RS, (c) the area of triangle PRS.
0
2
[ (a) 54.31 (b) 4.157 cm (c) 74.75 cm ]
(6) In Diagram 6, BCD is a straight line.
Diagram 6 Calculate (a) the length of AB, (b) CAD, (c) the area of triangle ACD.
0
2
[ (a) 6.678 cm (b)84.74 (c) 13.17 cm ]
Solutions of Triangles
168
Past year questions 1. SPM 2003 P2Q 15 Diagram below shows a tent VABC in the shape of a pyramid with triangle ABC as the horizontal base. V is the vertex of the tent and the angle between the inclined plane VBC and the base is 50 Given that VB =VC =2.2 m and AB =AC =2.6m, calculate (a) the length of BC if the area of the base is 3 m2. [3 marks] (b) the length of AV if the angle between AV and the base is 250. [3 marks] (c) the area of triangle VAB [4 marks] [ANSWERS; 2.700, 3.149, 2.829 ] 2. SPM 2004 P2 Q13 Diagram below shows a quadrilateral ABCD such that ABC is acute.
(a)Calculate
ABC (ii) ADC (i)
(iii) the area, in cm2, of quadrilateral ABCD [ 8 marks]
(b) A triangle A’B’C’ has the same measurements as those given for triangle ABC, that is, A’C’=12.3 cm, C’B’=9.5cm and B ' A 'C ' 40.5 , but which is different in shape to triangle ABC. (i) etch the triangle A’B’C’ (ii) State the size of A'B'C' [2 marks] [ANSWERS ; 57.23, 106.07, 80.96, 122.77] 3. SPM 2005 P2Q12 Diagram below shows triangle ABC (a) Calculate the length, in cm, of AC. [2 marks] (b) A quadrilateral ABCD is now formed so that AC is a diagonal, ACD 40 and AD =16 cm. Calculate the two possible values of ADC [2 marks] ( c) By using the acute ADC from (b) , calculate (i) the length , in cm, of CD (ii) (ii) the area, in cm2, of the quadrilateral ABCD [6 marks]
[ANSWERS; 19.27, 50.73, 24.89, 290.1 ]
Solutions of Triangles
169
4. SPM 2006 P2 Q 13 Diagram below shows a quadrilateral ABCD. The area of triangle BCD is 13 cm2 and BCD is acute. Calculate (a) BCD , [2 marks] (b) the length, in cm, of BD, [ 2 marks] (c) ABD , [3 marks] 2 (d) the area, in cm , quadrilateral ABCD [3 marks]
[ANSWERS ; 60.07, 5.573, 116.55 35.43 ]
5. SPM 2007 P2Q15 Diagram shows quadrilateral ABCD. (a) Calculate (i) the length, in cm, of AC (ii) ACB [4 marks] (b) Point A’ lies on AC such that A’B =AB. (i) Sketch A ' BC (ii) Calculate the area , in cm2 , of A ' BC [6marks]
[ANSWERS 13.36, 23.88 13.80] 6. SPM 2008 P2Q14 In the diagram below, ABC is a triangle. ADFB,AEC and BGC are straight lines. The straight line FG is perpendicular to BC.
It is given that BD = 19 cm, DA =16 cm, AE = 14 cm, DAE 80 and FBG 45 (a) calculate the length, in cm, of (i) DE (ii) EC (b) The area of triangle DAE is twice the area of triangle FBG, Calculate the length , in cm, of BG (c) Sketch triangle A’B’C’ which has a different shape from triangle ABC [ANSWERS: 19.344, 16.213, 10.502] Solutions of Triangles
170
[5 marks] [4 marks] [ 1 mark]
Solutions of Triangles
171
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