Similar Triangles, Right Triangles, And The Definition Of The Sine, Cosine And Tangent Functions Of Angles Of A Right Triangle Worksheets With Answer Key Page 9
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43Example: Solve ∆ABC where a = 22, c = 27 and A = 50°.
Solution: By the Law of Sines,
°
27
22
27
sin50
=
=
=
so
sin
C
. 0
9401
.
°
sin
C
sin
50
22
There are two angles C between 0° and 180° for which sin C = 0.9401. They are
C
= 70.07° and C
= 109.93°. Since
1
2
A + C
= 50° + 70.07° = 120.07° < 180°
1
and
A + C
= 50° + 109.93° = 159.93° < 180°,
2
C
and C
each lead to a solution of the triangle. There are two triangles with given
1
2
sides and angle. We must find them both.
When C = C
= 70.07°, B = 180° - (70.07° + 50°) = 59.93°. According to the Law of
1
Sines, then
°
b
22
22
sin59.93
=
=
=
so
b
24
.
85
.
°
°
°
sin
59
.
93
sin
50
sin50
The first solution to this triangle is
a = 22,
b = 24.85,
c = 27,
A = 50°, B = 59.93°, C = 70.07°.
When C = C
= 109.93°, B = 180° - (109.93° + 50°) = 20.07°.
2
Again by the Law of Sines,
°
b
22
22
sin20.07
=
=
=
so
b
. 9
86
.
°
°
°
sin
20
.
07
sin
50
sin50
The second solution to this triangle is
a = 22,
b = 9.86,
c = 27,
A = 50°, B = 20.07°, C = 109.93°.
Example: Solve ∆ABC where a = 13, c = 6 and A = 70°.
Solution: By the Law of Sines,
°
6
13
6
sin
70
=
=
=
so
sin
C
. 0
4337
.
°
sin
C
sin
70
13
There are two angles C between 0° and 180° for which sin C = 0.4337. They are
C
= 25.7° and C
= 154.3°. Since
1
2
A + C
= 70° + 25.7° = 95.7° < 180°,
1
C
can be an angle of a triangle which has the given sides and angle. Since
1
A + C
= 70° + 154.3° = 224.3° > 180°,
2
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