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Lesson 40 – Sine Law & The Ambiguous Case Math 2 Honors - Santowski 10/24/20151Math 2 Honors -...
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Transcript of Lesson 40 – Sine Law & The Ambiguous Case Math 2 Honors - Santowski 10/24/20151Math 2 Honors -...
Lesson 40 – Sine Law & The Ambiguous Case
Math 2 Honors - Santowski
04/20/23 1Math 2 Honors - Santowski
Lesson Objectives
Understand from a geometric perspective WHY the ambiguous case exists
Understand how to identify algebraically that their will be 2 solutions to a given sine law question
Solve the 2 triangles in the ambiguous case
See that the sine ratio of a acute angle is equivalent to the sine ratio of its supplement
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3
(A) Review of the Sine Law
If we have a non right triangle, we cannot use the primary trig ratios, so we must explore new trigonometric relationships.
One such relationship is called the Sine Law which states the following:
AB
C
sinsinsin
OR sinsinsin c
C
b
B
a
A
C
c
B
b
A
a
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4
Law of Sines: Solve for Sides
C
c
A
a
b
B
Have: two angles, one side opposite one of the given angles
Solve for: missing side opposite the other given angle
Missing Side
a sin A
= b sin B
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5
Law of Sines: Solve for Angles
C
c
A
a
b
B
Have: two sides and one of the opposite angles
Solve for: missing angle opposite the other given angle
Missing Angle a sin A
= b sin B
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6
(B) Examples Sine Law
We can use these new trigonometric relationships in solving for unknown sides and angles in acute triangles:
ex 4. Find A in ABC if a = 10.4, c = 12.8 and C = 75°
ex 5. Find a in ABC if A = 84°, B = 36°, and b = 3.9
ex 6. Solve EFG if E = 82°, e = 11.8, and F = 25°
There is one limitation on the Sine Law, in that it can only be applied if a side and its opposite angle is known. If not, the Sine Law cannot be used.
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7
(B) Examples Sine Law
Mark is a landscaper who is creating a triangular planting garden. The homeowner wants the garden to have two equal sides and contain an angle of 75°. Also, the longest side of the garden must be exactly 5 m. (a) How long is the plastic
edging that Mark needs to surround the garden?
(b) Determine the area of the garden.
5 meters
75
H
F
G
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(C) Inverse Trig – Input/Output Determine the measure of an angle whose sine ratio is 0.75 Solve the equation sin(x) = 0.75 for x Solve the equation x = sin-1(0.75)
What is the difference in meaning amongst these 3 questions??
Solve the following equations for x: x = sin-1(0.89) x = cos-1(.11) x = sin-1(0.25) x = tan-1(3.25) sin(x) = 0.45 sin(x) = 0.6787
Explain why it is IMPOSSIBLE to solve sin-1(1.25) = x
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(D) Sine Law – Scenario #1
Let’s work through 2 scenarios of solving for B :
Let A = 30°, a = 3 and b = 2 (so the longer of the two given sides is opposite the given angle)
Then sin = b sin / a And sin = 2 sin 30 / 3 So B = 19.5°
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(D) Sine Law – Scenario #2
In our second look, let’s change the measures of a and b, so that a = 2 and b = 3 (so now the shorter of the two given sides is opposite the given angle)
Then sin = b sin / a And sin = 3 sin 30 / 2 So B = 48.6°
BUT!!!!! …….. there is a minor problem here …..
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(E) Altitude (height)
Explain how to find the height (altitude) of this non-right triangle
a = 22.8 cm
m1 = 46
B
C
A
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(E) Altitude (height)
side of unknown length ==> X
bsin A = 14.7 cm
b = 23.6 cm
altitude = 14.7 cm
A = 38
A B
C
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(F) Considerations with Sine Law If you are given information about non-right triangle and you
know 2 angles and 1 side, then ONLY one triangle is possible and we never worry in these cases
If you know 2 sides and 1 angle, then we have to consider this “ambiguous” case issue If the side opposite the given angle IS THE LARGER of the 2
sides NO WORRIES If the side opposite the given angle IS THE SHORTER of the
2 sides ONLY NOW WILL WE CONSIDER THIS “ambiguous” case
WHY????
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Case #1 – if a>b
side of unknown length ==> X
side a
If: (i) a > b, then ONE OBTUSE triangle is possible
altitude = 11.3 cm
a = 20.2 cm
A = 45
b = 16.0 cm
A B
C
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Case #2
side of unknown length ==> X
side a
If: (i) a < b AND (ii) b sinA (altitude) > a, then NO triangle is possible
altitude = 11.4 cm
a = 8.6 cm
A = 45
b = 16.0 cm
A B
C
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Case #3
side of unknown length ==> X
side a
If: (i) a < b AND (ii) b sinA (altitude) = a, then ONE RIGHT triangle is possible
altitude = 11.4 cm
a = 11.4 cm
A = 45
b = 16.0 cm
A B
C
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Case #4 – the Ambiguous Case
side of unknown length ==> X
sin mCBA = 0.83
sin mB'BC = 0.83
sin mCB'B = 0.83
mB'BC+mCBA = 180.00
mB'BC = 55.72
mCBA = 124.28
mCB'B = 55.72
a = 10.2 cm
If: (i) a < b, AND (ii) if bsinA < athen TWO triangles are possible
altitude = 8.5 cm
A = 36
b = 14.4 cm
B'E
B
04/20/23 17Math 2 Honors - Santowski
Case #4 – the Ambiguous Case
side of unknown length ==> X
sin mCBA = 0.83
sin mB'BC = 0.83
sin mCB'B = 0.83
mB'BC+mCBA = 180.00
mB'BC = 55.72
mCBA = 124.28
mCB'B = 55.72
a = 10.2 cm
If: (i) a < b, AND (ii) if bsinA < athen TWO triangles are possible
altitude = 8.5 cm
A = 36
b = 14.4 cm
B'E
B
04/20/23 18Math 2 Honors - Santowski
Case #4 – the Ambiguous Case
side of unknown length ==> X
sin mCBA = 0.83
sin mB'BC = 0.83
sin mCB'B = 0.83
mB'BC+mCBA = 180.00
mB'BC = 55.72
mCBA = 124.28
mCB'B = 55.72
a = 10.2 cm
If: (i) a < b, AND (ii) if bsinA < athen TWO triangles are possible
altitude = 8.5 cm
A = 36
b = 14.4 cm
B'E
B
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Summary
Case 1 if we are given 2 angles and one side proceed using sine law
Case 2 if we are given 1 angle and 2 sides and the side opposite the given angle is LONGER proceed using sine law
if we are given 1 angle and 2 sides and the side opposite the given angle is SHORTER proceed with the following “check list”
Case 3 if the product of “bsinA > a”, NO triangle possible Case 4 if the product of “bsinA = a”, ONE triangle Case 5 if the product of “bsinA < a” TWO triangles
RECALL that “bsinA” represents the altitude of the triangle
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Summary
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Examples of Sine Law
if A = 44∠ º and B = 65∠ º and b=7.7 find the missing information.
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Examples of Sine Law
if A =44.3∠ º and a=11.5 and b=7.7 find the missing information.
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Examples of Sine Law
if A =44.3 and a=11.5 and b=7.7 find the ∠missing information.
04/20/23 24Math 2 Honors - Santowski
Examples of Sine Law
if A=29.3∠ º and a=12.8 and b = 20.5
04/20/23 25Math 2 Honors - Santowski
Examples of Sine Law
if A=29.3 and a=12.8 and b = 20.5∠ All the other cases fail, because bsinA<a<b
10<a (12.8)<20.5, which is true. Then we have two triangles, solve for both
angles
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Examples of Sine Law
Solve triangle PQR in which ∠ P = 63.5° and ∠ Q = 51.2° and r = 6.3 cm.
04/20/23 27Math 2 Honors - Santowski
Examples of Sine Law
Solve triangle PQR in which ∠ P = 63.5° and ∠ Q = 51.2° and r = 6.3 cm.
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Examples of Sine Law
ex. 1. In ΔABC, ∠ A = 42°, a = 10.2 cm and b = 8.5 cm, find the other angles
ex. 2. Solve ΔABC if ∠ A = 37.7, a = 30 cm, b = 42 cm
04/20/23 29Math 2 Honors - Santowski
Examples of Sine Law
ex. 1. In Δ ABC, ∠ A = 42°, a = 10.2 cm and b = 8.5 cm, find the other angles
First test side opposite the given angle is longer, so no need to consider the ambiguous case i.e. a > b therefore only one solution
ex. 2. Solve Δ ABC if ∠ A = 37.7, a = 30 cm, b = 42 cm First test side opposite the given angle is shorter, so we need to
consider the possibility of the “ambiguous case” a < b so there are either 0,1,2 possibilities.
So second test is a calculation Here a (30) > b sin A (25.66), so there are two cases
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Homework
HW
Nelson Questions: any of Q5,6,8
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