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TRIANGLES
Solutions of Oblique Triangles
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Conditions that Determine a Triangle
A triangle is said to be determined whenthe measures of three parts are given.
Three angles do not serve to determinea triangle. A triangle is determined inany of the following cases:
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Cases:Case 1. Law of Sines.
Given two angles and one side. Obtain the thirdangle, and then use the sine law.
Case 2. Ambiguous Case.
Given two sides and the angle opposite one of
them. Use the sine law. Examine the possibility ofno solution, one or two solutions.
Case 3. Given two sides and the included angle.
Use the cosine law for the third side and the sine
law for the angles. Find the smaller angle first toavoid the possibility of an obtuse angle.
Case 4. Given three sides,
Use the cosine law for each angle in turn.
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Oblique Triangles
c = 20
b a
A
C
B
0
300
65
1.
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2.
b = 24 a = 24
C = 18A B
C
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Oblique Triangles
An oblique triangle is a triangle which doesnot contain a right angle of 900. It contains
either three acute angles, or two acuteangles and one obtuse angle.
All acute angles One obtuse angle
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A triangle is uniquelydetermined when three parts,
not angles are known. Thusany triangle problem may fallunder any one of the
following cases:
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CasesCase I. Given one sideand two
angles
Case II. Given two sidesand an
angleopposite one of them
Case III. Given two sidesand theincluded angle.
Case IV. Given the three sides
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Law of Sines
In any triangle, the sides areproportional to the sines of the
opposite angles, that is,a b c
-------- = -------- = --------
sin sin sin
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Note:
We say,
sin a sin b sin c
------ =----; -------- =----; ------- =----
sin b sin c sin a
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Proof ( The given angle is acute )
Let ABC be any oblique triangle, In the figureangles and are acute. Let CDAB anddenote h = CD.
A B
C
D
hb
In the right triangle ACD above,h = b sin
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Proof ( The given angle is obtuse )
Let ABC be any oblique triangle,In the figure angleis obtuse. Let CDAB and denote h = CD.
A B
C
D
In the right triangle ACD above,h = a sin
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Thus,
h = h
b sin = a sin
sin a
---------- = ----------
sin b
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Similarly,
we obtain
a c b c
------- = -------- or --------- = --------
sin sin sin sin
a b c
---------- = ----------- = -------------
sin sin sin
Therefore,
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Cont.
Solutions ofOblique Triangles
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Case I
Given one side and two angles
Let a, , be given.
To find , use= 1800 ( + ).
To find b, use
sin b
------- = -------
sin a
or
a sin
b = --------
sin
To find c, use
sin c
------- = -------
sin a
or
a sin
c = -------
sin
a
b
c
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Example 1Solve ABC, given c = 20, = 300, = 650
c = 20300 650
A B
C
= 1800 - ( 300 + 650 ) = 850
ac sin
a = ------------
sin
20 sin 300
= ----------------
sin 850
= 10.4
c sin
b = ------------
sin
20 sin 650
= ------------
sin 850
= 18.20
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Example 2
A and B are two points on opposite bankof a river. From A a line AC = 270 ft. islaid off and the angles CAB = 1200 and
ACB = 450 are measured. Find thelength of AB.
A
B
A
C
b = 270 ft.1200
450
c = ?
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Solution
= 1800 ( 1200 + 450 ) = 150
b sin
c = ------------
sin
270 sin 450
= -------------
sin 15
737.65 ft.=
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Case II
Given two sides and the angle opposite one of
them. Let b, c and are given.
From
sin c
-------- = -----,
sin b
c sin
sin = ----------
b
If sin > 1 no angle is determinedIf sin = 1, = 900, a right triangle is determined
If sin < 1, two angles are determined:a.) an acute angle and
b.)an obtuse angle = 1800 -
Thus, there may be one or two triangles determined.
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Geometrically,
Let us discuss several special cases ofCase II, given two sides
angle opposite one of them.
a b ba
Let a, b, and be the given parts.
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ab
a
c
Given two sides and the angle opposite one of them.
Examples:
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The given angleis acute (1st)
In the figure: when b < AD and AD= c sin ,then the arc does not meet BE, hence notriangle is determined.
bc
A
DB E
c > b
h > b
h Thus, nois formed
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The given angleis acute (2nd)
In the figure, when b = AD, then the arcis tangent to BE, hence one right
triangle is determined.
B D
A
c b
E
c > b
c > h
b = h, h = c sin
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The given angleis acute (3rd)
In the figure, when b > AD but b < c, thenthe arc meets BE in two points C and C.Hence, two triangles ABCand ABC aredetermined. In ABC, is an acute anglewhile in ABC, = 1800 is an obtuseangle.
b
A
B C
c
DC
c > bc > b > h
hb
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The given angleis acute (4th)
In the figure, when b > AD and b = c, then
the arc meets BE in C and B. Hence oneisosceles triangle is determined.
bc = b
A
B C
D E
c = bb > h
h
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The given angleis acute (5th)
In the figure, when b > c, then the arcmeets BE in C and BE extended in C.Since ABC does not contain the given
angle , only one triangle is determined.
A
B C
bc
a E
C
b > c
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The given angle is obtuse(1st)
In the figure, when b < c or b = c, thenno triangle is determined.
A
B E
c
b b < c
b = c
Thus, nois formed
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The given angle is obtuse(2nd)
In the figure, when b > c, then only onetriangle is determined
A
B C
c
a
b
E
The arc intersects BE in two points but there is only
one which determines a triangles containing.
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1. When the angle is acute, there will be
a.) one solutionif the side opposite thegiven angle is greater than or equal to
the other given side, and
b.) no solution, one solution or
two solutionsi f the side opposi te the
given angle is less than the other given s ide.
c sin
sin = -------------
b
This sub case is determ ined by comput ing
To summarize this ambiguous case:
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2. When the angle is acute, there will be
b.) one solutionif the side opposite thegiven angle is greater than the other given
side.
a.) no solutionif the side opposite the givenangle is less than or equal to the other given side.
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Example 3
a.) When b = 20, c = 10, = 300
there is one solution since isacute and b > c.
Answer:
b = 20c = 10300
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b.)When b = 10, c = 20, = 300
Answer:
there is no solution, one solution or
two solution.
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c.)When b = 25, c = 15, = 1300
Draw the figure:
Answer:
there is one solution.
b = 25
c = 15 1300
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d.)When b = 25, c = 35, = 1300
Answer:
there is no solution
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Example 4
Solutions:
Since is acute and c > b,
there is only one solution.
For :
b sin
s in = --------------
c
80 sin 620
= ------------------
128
= 0.5518
s in = 0.5518 ,
=
330
30
620b = 80
c =128
Solve ABC, given c = 128, b = 80, = 620.
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For: = 1800( + )
= 84030
b sin 80 sin 840
30
For a:a = --------------- = -------------------- =144.28
sin sin 33030
Example 4 (solution)
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Example 5
Solve ABC, given a = 425, c = 310, = 120050
120050
A
B
C
a = 425c = 310
Example 5 (solution)
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Example 5 (solution)Solve ABC, given a = 425, c = 310, = 120050
Solution:
Since is obtuse and a > c, there is one solution.
c sin 310 sin 120050
For : sin = ----------- = -----------------------, =38046
a 425
For : = 1800 ( + ) = 20024
a sin 425 sin 20024
For b: b = -------------- = -------------------- = 172.53
sin sin 120050
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Law of CosinesIn any ABC, the square of any side isequal to the sum of the squares of theother two sides minus twice the product of
the sides and the cosine of their includedangle, that is,
a2= b2+c22bc cos ,
b
2
= a
2
+c
2
2ac cos c2= a2+ b22ab cos .
C
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Proof1:
In the right triangleACD in the figure
In the right BCD in the figure,h = a sin and DB = a cos .
A
C
BD
bh
b2 = h2 + AD2
a
AD = ABDBThen,
c
= c - a cos
b2= a2sin2+ ( c a cos )2
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So,
b2=a2sin2+ c22 ac cos +a2cos2
Therefore,
b2= a2+ c22 ac cos .
b2
=a2
sin2
+ a2
cos2
+ c2
2 ac cosb2=a2(sin2+ cos2)+ c22 ac cos
b2= a2sin2+ ( c a cos )2
C
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Proof2:
In the right triangleACD in the figure
A
C
B
b2 = h2 + AD2
D
bh
In the right BCD in the figure,
a
h = a sin = a sin (1800-)
= a sin
BD = a cos
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Therefore:
b2 = h2 + AD2A D
bh
= a2sin2 + (c a cos )2
= a2sin2 + c22ac cos +a2cos2
= a2(sin2 + cos2)+ c22ac cos
b2 = a2 + c2 2ac cos
Note:The other formulas are obtainedby cyclic changes of the letters.
B
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CASE III
Given two sides and the included angleLet a, b, and are given.
To find c: use c2= a2+ b22ab cos .
To find : use
use
a sin sin = -------------
c
To find :
b sin sin = -------------.
c
To check, use + + = 1800.
B
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Example 6.
Solve ABC, given a = 150, b = 230, = 32020.
For c: c2= a2+ b22ab cos
= 1502+ 23022 (150)(230) cos 32020
c
2
= 17098.39c = 130.76a sin 150 sin 32020
For : sin =-----------=-------------------; = 37050c 130.76
For : = 1800( + )= 1800( 37050 + 32020)= 109050
A
B
C32020
a = 150
b = 230
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Example 7Two forces 20.5 kgs and 25.5 kgs act on
a body. If their directions make anangle of 60020 with each other, findthe magnitude of their resultant and the
angle which it makes with the largerforce.
20.5
25.5
6002020.5
25.5
A B
CD
C25.5D
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Solution:
In parallelogram ABCD,
A + B = C + D = 1800
20.5
25.560
0
20
20.5
A B
B= 180060020 =119040
In ABC, b2= a2 + c2 2 a c cos b2= 20.52+ 25.522(20.5)(25.5) cos 119040
b =39.85kgs. (magnitude)a sin 20.5 sin 119040
sin = ------------- = ----------------------, =26033b 39.85
a
c
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CASE IV
Given three sides
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Let a, b and c be given. Apply the law of cosines foreach of the angles. To find the angles
a2+ b2c2
cos =---------------------2ab
b2+ c2a2
cos = -----------------,2bc
a2
+ c2
b2
cos =-------------------,2ac
To check, use + + = 1800.
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Example 8.
Solve ABC, given a=26.4, b = 34.5, c = 52.8.
a=26.4b = 34.5
c = 52.8
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Solution:
b2+ c2- a2 34.52+ 52.8226.42
For : cos = ----------------- = ------------------------- ; = 250452bc 2(34.5)(52.8)
a2+ c2b2 26.42+ 52.8234.52
For : cos = ---------------- = --------------------------; = 340362ac 2 (26.4) (52.8)
a2
+ b2
c2
26.42
+ 34.52
52.82
For : cos = ----------------- = -------------------------------; = 119038
2ab 2(26.4)(34.5)
a=26.4b = 34.5
c = 52.8
Check: + + = 1800
E l 9
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Example 9.
An isosceles triangle has sides thatmeasure 24 cm., 24 cm and 18 cm.Find the measure of each angle.
a = 24b = 24
c = 18
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Solution:242+ 182242
cos = --------------------------- , =670582(24)(18)
182+ 242242cos = -------------------------, 67058
2 (18)(24)
242+ 242 - 182cos = -------------------------, = 44003
2 (24)(24)
Check: + + = 1800
a=24b=24
c = 18
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Thank Youand
Mabuhay!