Transcript 9.3 The Law of Sines

FOUR
CASES
CASE 1: One side and two angles are
known (SAA or ASA).
CASE 2: Two sides and the angle
opposite one of them are known (SSA).
CASE 3: Two sides and the included
angle are known (SAS).
CASE 4: Three sides are known (SSS).
A
S
A
ASA
S
A
A
SAA
CASE 1: ASA or SAA
S
A
S
CASE 2: SSA
S
A
S
CASE 3: SAS
S
S
S
CASE 4: SSS
The Law of Sines is used to
solve triangles in which Case 1
or 2 holds. That is, the Law of
Sines is used to solve SAA,
ASA or SSA triangles.
In A B C :
sin A

a
sin B

sin C
b
c

 This is used in triangles that are not right triangles.
If we know 2 angles and 1 side of the triangle then
we can use the law of sines. For example…
a
25
20
o
b
o
sin 2 5
8 0 .4

sin 20
a
 The last example shows a case of SAA or AAS
 Or ASA.

a
25
20
o
o
b
With that set up of ASA or AAS there is one and only
one triangle that can be formed.
The
AMBIGUOUS
Case
Open to or having several possible meanings or interpretations.
Of doubtful or uncertain nature; difficult to comprehend,
distinguish, or classify
Lacking clearness or definiteness; obscure; indistinct
If given SSA there are several situations that can
happen.
1 POSSIBILITY C
C
b
NO
a POSSIBILITIES
b
A
A
C
C
2 POSSIBILITIES
b
A
a
1 POSSIBILITY
b
a
A
a
•General Steps for solving the ambiguous case You
will be given two sides and 1 angle. Use these and the
formula for the law of sines to get the second angle
•Check if the first angle that you found is valid
• i.e. Do the two angles that you currently have
have a sum that is less than 180
•Check if there is a second angle that is valid. To do
this, find the angle in Quadrant II that has the same
sine as the angle that you found in Step 2 (subtract that
angle from 180)
•If the angle in Quadrant II plus the original angle you
were given have a sum that is less than 180, you have a
second triangle that works.
If the given information presents you with a triangle of the
orientation SSA, one side can be drawn in two different
positions as illustrated below. Because of this we will
check for 2 possible triangles.
Solve ∆ABC if  A  43.1 , a  186.2, and b  248.6
o
C
b=248.6
43.1°
A
a=186.2
a=186.2
B2
B1
b=248.6
a=186.2
43.1°
a=186.2
A
B1
B2
From the Law of Sines we can find sinB
sin B
b

sin A
sin B
a
248.6

sin 43.1
186.2
sin B = 0.91225
sin B 
248.6 sin 43.1
186.2
sin B = 0.91225
Using sin  1 we can find the measure of angle B1,
which is approximately 65.82°.
To check to see if this angle is valid we start with
the angle that was given (43.1° and add 65.82° to
ensure that the sum is less than 180°)
43.1°+65.82°=108.92°
Thus making the third angle approximately 71.08°.
But what about the possibility of a second triangle
with angle at B2 because we started with SSA?
Take the 65.82° that we found and find its
supplement: 180°-65.82°=114.18° (this is done because
the sine of an angle is equal to the sine of its
supplement, this is due to the fact that sine is positive
in quadrants I and II, in hand with the ideas of
reference angles.)
Now check to see if that angle and the angle we
started with of 43.1° has a sum of less than 180°.
114.18° + 43.1° = 157.28°, so with this angle another
triangle is possible and its third angle would be
22.72°
If either of the angles that I found in the
previous example would have been added
to 43.1° and the sum of the two angles was
greater than 180 °, then that angle would
not be a valid angle therefore that triangle
would not be able to exist in the geometry
that we know (Euclidian), but I would still
want to check for a second possibility.
A tree on a hillside casts a
shadow 209 ft down the hill.
If the angle of inclination of
the hillside is 21 degrees to the
horizontal and the angle of
elevation of the sun is 58
degrees, find the height of the
tree.
58°
21°
Homework
