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Warm up A 5.2 m ladder leans against a wall. The
bottom of the ladder is 1.9 m from the wall. What angle does the ladder make with the ground (to the nearest degree)?
Cos-1(1.9/5.2) = 69o
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Objective: To use the Law of Sines in order to solve oblique triangles
The Law of Sines
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Consider the first category, an acute triangle (, , are acute).
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hNow, sin( ) , so that h a sin( )
ah
But sin( ) , so that h c sin( )c
By transitivity, a sin( ) c sin( )
sin( ) sin( )Which means
c a
Create an altitude, h.
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Theorem Law of Sines
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The Law of Sines is used when we know any two angles and one side or when we know two sides and an angle opposite one of those sides.
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Applying the Law of Sines The Law of Sines may be used when the
known parts of the triangle are:◦ 1. one side and two angles (SAA), (ASA)◦ 2. two sides and an angle opposite one of the
sides (SSA)
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Example: In triangle ABC, angle A = 106 o, angle B =
31o and side a = 10 cm. Solve the triangle ABC by finding angle C and sides b and c.(round answers to 1 decimal place).
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Use the fact that the sum of all three angles of a triangle is equal to 180 o to write an equation in C. A + B + C = 180 o Solve for C. C = 180 o - (A + B) = 43 o
Solution
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Use sine law to write an equation in b. a / sin(A) = b / sin(B) Solve for b. b = a sin (B) / sin(A) = (approximately) 5.4 cm
Use the sine law to write an equation in c.a / sin(A) = c / sin(C)Solve for c. c = a sin (C) / sin(A) = (approximately) 7.1 cm
Solution (cont’d)
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Area of an Oblique Triangle
The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle.
Referring to the triangles below, that each triangle has a height of h = b sin A.
A is acute. A is obtuse.
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Area of a Triangle - SASSAS – you know two sides: b, c and
the angle between: A
Remember area of a triangle is ½ base ● height
Base = bHeight = c ● sin A Area K= ½ bc(sinA)
A
B
C
c a
b
h
Looking at this from all three sides:K = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)
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Example: Find the area ofgiven a = 32 m, b = 9 m, and
ABC36 .m C
132 9 sin36
2Area m m
284.6Area m
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You can also find the area if you know one side and 2 angles based on the Law of Sines.
so, substitute for b in the last
equation, K = ½ bc(sinA) gives you
Area of a Triangle
C
c
BSin
b
sin
C
Bcb
sin
sin
C
BAcK
sin
sinsin
2
1 2
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Find the area of triangle JKL if j=45.7, K=111.1o, and L=27.3o.
Area of a Triangle
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http://www.emathematics.net/trigonometria.php?tr=5
Law of Sines practice
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Using the triangle above, A = 50o, B = 65o and a = 12. Solve the triangle.
Warm up
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Lesson 5-7 Law of Sines the Ambiguous Case
Objective: To determine whether a triangle has zero, one or two solutions and solve
using the Law of Sines.
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The Ambiguous Case – SSA
In this case, you may have information that results in one triangle, two triangles, or no triangles.
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SSA – No Solution
Two sides and an angle opposite one of the sides.
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By the law of sines,
sin(57 ) sin( )
15 20
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Thus,20 sin(57 )
sin( )15
20 (0.8387)sin( )
15sin( ) 1.1183 Impossible!
Therefore, there is no value for that exists! No Solution!
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SSA – Two Solutions
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By the law of sines,
sin(32 ) sin( )
30 42
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So that,
42 sin(32 )sin( )
3042 (0.5299)
sin( )30
sin( ) 0.7419
48 or 132
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Case 1 Case 2
48 32 180
100
132 32 180
16
Both triangles are valid! Therefore, we have two solutions.
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Case 1 )sin(100 sin(32 )
c 30
30 sin(100 )c
sin(32 )
30 (0.9848)c
0.5299
c 55.7539
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Case 2 sin(16 ) sin(32 )
c 30
30 sin(16 )c
sin(32 )
30 (0.2756)c
0.5299
c 15.6029
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For SSA Triangles:
1. If A < 90°a. a < b
1. a < b(sin A) No Solution2. a = b(sin A) 1 Solution3. a > b(sin A) 2 Solution
b. a ≥ b 1 Solution
2. If A ≥ 90°a. a ≤ b No Solution
b. a > b 1 Solution
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Practice Solve the triangle: A = 42°, a = 11, and b =
6