area of oblique triangles
TRANSCRIPT
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CALCULUS
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Oblique Triangles
Oblique triangles do not have any 90oangles
45o
72o
63o
120o
24o
46o
A
B
C
A
B
C
a
b
c
a
bc
Acute angled
Obtuse angled
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METHOD 1A
B D C
If we want to find the
area of a but the height
(altitude) is unknown
Stage 1
(small )
In ACD a
bh
h = b sinc
hin
Stage 2
(big )
In ABC
A = (base )(height)
But we just determined height
from using a smaller part of
the triangle
Substitute height
A = ( a )( b sinc )
A = a b sinc
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This is used to find the area of oblique triangles if
you know two sides and the included angle
A
CB
48o
25 cm
c
b
a
A = a c sinB
A= (30cm)(25 cm) sin 48o
A= (30cm)(25 cm) (0.743)
A= (30cm)(25 cm) (0.743)
A= 278.625 cm2
EXAMPLE:
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METHOD 2
When the two angles and a side is known. The area of the triangle
is equal to the square of one side and twice the sine of the oppositeangle multiplied by the sine of the other angles.
FORMULA:
Given side A and the other three angles
Area =
Given side B and the other three angles
Area =
Given side C and the other three angles
Area =
2 sin
sin
2sin
2 sin sin2sin
2 sin sin2sin
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EXAMPLES
1. B SOLUTION:Area = 2 sin sin
2sin 522
=(6.9 )2 sin 8048
2sin 52 = 13.73
A C
2. E SOLUTION:
Area = 2 sin sin 2 sin
Y =(7.5 )2 sin3075
2 sin 75 = 13.12
S
3. O SOLUTION:
Area =2 sin sin
sin
=(7.5)2 sin3075
2 sin 75 = 9.32
D G
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METHOD 3 If all three sides are known, the area can be determined using
HERONS Formula.
FORMULA:
Lets represent half of the perimeter of a triangle and a, b and c
as its sides then,
Where: s =
The area of herons formula is stated as:
Area =
++2 : ++2
( )
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Case 2 Find the area of a triangle given 3 sides
24 m
20 m10 m
Herons Formula
Step 1: Find the semi-perimeter
2
cbas
2
cbas
2102420 mmms
ms 27
Step 2: area formula
))()(( csbsassA ))()(( csbsassA
)1027)(2027)(2427(27 A
)17)(7)(3(27A
9639A 22.98 mA
B
C A
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2
cbas
))()(( csbsassA
Herons Formula
52 m
65 m
28 m
B
C
A
2
655228 mmms
ms 5.72
)5.7)(5.20)(5.44(5.72A
9375.496035A
2704.298mA
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METHOD 4
When the vertices are given, this method is called the
Matrix = adcd
Using the determinants, we have the two formulas:
METHOD A
1 1 12 2 13 3 1= 1(23)- 1(2 3)+ 1(23 (32)
The area of this would be : Area =1
2
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METHOD B:
1 1 12 2 1
3
3 1
1 1 12 2 1
3
3 1
= 12+13+23 32+13+12
Area =
1
2
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1. (3,4) SOLUTION:Area =
1
2
Q = 12 21A = 10 5 u2
(6, -2) (8, 1)
3 4 16 2 18 1 1
D = x1(y2
y3)
y1(x2
x3) + 1(x2y3
y3x3)= 3(-2 - 1) 4(6-8) + 1(6 (-2)(8))
= -9 + 8 +22
D = 21
EXAMPLES
1. (4,6) SOLUTION:Area =
1
2 =
1
2
A = 2u2(9,3)
(7,5)
4 6 19 3 17 5 1
D = 4(3 - 5) 6(9 - 7) + 1(9)(5) (3)(7)
= -8 12 + 24
D = 4
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END