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Solutions to various triangle problems using the law of cosines and law of sines. It includes step-by-step calculations to find the missing sides and angles of triangles given certain information.
Typology: Assignments
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Math 112 Jonny Comes
Winter 2007 Solutions to additional problems for assignment #
From the Textbook:
Solutions:
o
We want to find a. Using law of cosines we have
a
2 = (60.5)
2
2 − 2(60.5)(90) cos(
◦ )
Since a is positive we see a =
2 ft.
o
We want to find a. Using law of cosines we have
a^2 = (400)^2 + 90^2 − 2(400)(90) cos(45◦)
Since a is positive we see a =
2 ft.
[Notice the triangle inequality holds!] Using the law of cosines we see
cos(A) =
Thus A = arccos(
941 1060 ). Similarly
cos(B) =
Thus B = arccos(
44 159 ). Similarly
cos(C) =
Thus C = arccos( 1760977 ).
o
Additional Exercises: (Be sure to justify all your answers)
For the following problems use the triangle
A
B
C a
b
c
(i) a = 2, b = 2, and c = 2
Solution: We need to find A, B, and C. First let’s find A.
a
2 = b
2 +c
2 − 2 bc cos A ⇒ 2
2 = 2
2 +(
2 − 2 · 2 ·(
3)·cos A
3 cos A ⇒ −12 = − 8
3 cos A
⇒ cos A =
Since we know 0◦^ < A < 180 ◦^ we have
A = arccos
Now let’s find B.
b^2 = a^2 +c^2 − 2 ac cos B ⇒ 22 = 2^2 +(
3)·cos B
3 cos B ⇒ −12 = − 8
3 cos B
⇒ cos B =
Since we know 0◦^ < B < 180 ◦^ we have
B = arccos
◦ .
[We just did the same exact calculation as we did to find A. Maybe it would have been better to say b = a so by symmetry
B = A = 30◦]
To find C we can use the fact that A + B + C = 180◦^ to get
◦ − A − B = 180
◦ − 30
◦ − 30
◦ = 120
◦ .
(ii) a = 5, b = 7, and c = 6.
Solution: We need to find A, B, and C. First let’s find A.
a
2 = b
2
2 − 2 bc cos A ⇒ 5
2 = 7
2
2 − 2 · 7 · 6 · cos A
⇒ 25 = 49 + 36 − 84 cos A ⇒ −60 = − 84
3 cos A
⇒ cos A =
Since we know 0◦^ < A < 180 ◦^ we have
A = arccos
Now let’s find B.
b^2 = a^2 + c^2 − 2 ac cos B ⇒ 72 = 5^2 + 6^2 − 2 · 5 · 6 · cos B
⇒ 49 = 61 − 60 cos B ⇒ −12 = −60 cos B
⇒ cos B =
Since we know 0◦^ < B < 180 ◦^ we have
B = arccos
To find C we can use the fact that A + B + C = 180◦^ to get
◦ − A − B = 180
◦ − arccos
− arccos
(iii) a = 2, b = 6, and c = 3.
Solution: Well, a + c = 2 + 3 = 5 < 6 = b so the values given do not satisfy the Triangle Inequality. Thus there are no solutions.
(iv) a = 6, b = 3
2, and C = 45◦.
Solution: We need to find A, B, and c. First let’s find c.
c
2 = a
2
2 − 2 ab cos C = 6
2
2 − 2 · 6 · (
◦
⇒ c^2 = 36+18− 36
Since we know c is positive
c =
The shortest way to proceed from here is to notice that c = b so
we know B = C = 45◦^ and thus
◦ − B − C = 180
◦ − 45
◦ − 45
◦ = 90
◦ .
Now to find cos(15◦) we use the half angle id to get
cos(15◦) = cos
1 + cos(30◦)
2
√ 3 2
2+
√ 3 2
Since cos(15◦) is positive we have cos(15◦) =
2+
√ 3
a
2 = 41 − 40
Since we know a is positive we have
a =
Solution: To find C we use
To find b we use the law of sines to get
b
sin B
a
sin A
b
sin 30◦^
sin 45◦
⇒ b =
12 sin 30◦
sin 45◦^
2
2 2
2 2
Solution: To find A we use
◦ − B − C = 180
◦ − 120
◦ − 45
◦ = 15
◦ .
To find b we use the law of sines to get
b
sin B
c
sin C
b
sin 120◦^
sin 45◦
⇒ b =
3 sin 120◦
sin 45◦^
3 2
2 2
Solution: To find C we use
◦ − A − B = 180
◦ − 30
◦ − 30
◦ = 120
◦ .
To find b we use the law of sines to get
b
sin B
a
sin A
b
sin 30◦^
sin 30◦
⇒ b =
7 sin 30◦
sin 30◦^
Solution: To find A we use
◦ − B − C = 180
◦ − 135
◦ − 30
◦ = 15
◦ .
To find c we use the law of sines to get
c
sin C
b
sin C
c
sin 30◦^
sin 135◦
⇒ c =
4 sin 30◦
sin 135◦^
1 2
2 2
Solution: To find B we use
To find c we use the law of sines to get
c
sin C
a
sin A
c
sin 30◦^
sin 120◦
⇒ c =
6 sin 30◦
sin 120◦^
1 2
3 2
Note: For problems 11-17, there may be 0, 1, or 2 solutions.
So we have sin B = 1, and we know 0 ≤ B ≤ 180 ◦^ so either
Case 1 : B = arcsin (1) = 90◦
or
Case 2 : B = 180
◦ − arcsin (1) = 180
◦ − 90
◦ = 90
◦ .
Both cases are the same and we have:
2, and B = 135◦^ find A and C.
Solution: First we use the law of sines to find A:
sin A
a
sin B
b
⇒ sin A =
a sin B
b
4 sin 135◦ √ 2
2 2
So we have sin A = 2, but this is impossible because sin x cannot be
more than 1. Thus there are no solutions.
3, c =
2, and B = 60◦^ find A and C.
Solution: First we use the law of sines to find C:
sin C
c
sin B
b
⇒ sin C =
c sin B
b
2 sin 60◦ √ 3
3 2
So we have sin C =
√ 2 2 , and we know 0^ ≤^ C^ ≤^180
◦ (^) so either
Case 1 : C = arcsin
or
Case 2 : C = 180
◦ − arcsin
◦ − 45
◦ = 135
◦ .
Case 1: If C = 45◦^ then (using A + B + C = 180◦) we know
◦ − 60
◦ − 45
◦ = 75
◦ .
Case 2: C = 135◦^ then (using A + B + C = 180◦) we know
◦ − 60
◦ − 135
◦ = − 15
◦ .
But this is not possible.
So the final solution is
C = 45◦^ and A = 75◦
√ 2 2 ,^ c^ = 1, and^ C^ = 45
◦ (^) find A and C.
Solution: C is given to be 45◦, so all we need to do is find A. First we use the law of sines to find B:
sin B
b
sin C
c
⇒ sin B =
b sin C
c
2 2
sin 45◦
So we have sin B = 12 , and we know 0 ≤ B ≤ 180 ◦^ so either
Case 1 : B = arcsin
or
Case 2 : B = 180
◦ − arcsin
◦ − 30
◦ = 150
◦ .
Case 1: If B = 30◦^ then (using A + B + C = 180◦) we know
Case 2: B = 150◦^ then (using A + B + C = 180◦) we know
But this is not possible.
So the final solution is
A = 105
◦
2, c = 17, and C = 30◦^ find A and B.
Solution: First we use the law of sines to find A:
sin A
a
sin C
c
⇒ sin A =
a sin C
c
2 sin 30◦
17
(c) It is possible for the interior angles of a triangle to have measures 63 ◦, 74◦, and 42◦.
Solution: FALSE. We know the interior angle of a triangle must
sum to 180◦^ and
find all the roots of f (x).
Solution: Well
x^2 − 3 x + 2
x^2 + 3
x^4 − 3 x^3 + 5x^2 − 9 x + 6 − x^4 − 3 x^2
− 3 x^3 + 2x^2 − 9 x 3 x^3 + 9x
2 x^2 + 6 − 2 x^2 − 6
0
so f (x) = (x^2 + 3)(x^2 − 3 x + 2). So the roots of f (x) are the roots of
x^2 + 3 and the roots of x^2 − 3 x + 2.
The roots of x^2 + 3 are found by setting x^2 + 3 = 0 which implies
x^2 = −3 or x = ± 3 i.
The roots of x^2 − 3 x + 2 = (x − 2)(x − 1) are 2 and 1.
Thus the roots of f (x) are 1, 2 , 3 i, and − 3 i.
and f (2 − 3 i) = 0
(a) Write f (x) as a product of linear factors (there should be 6 of
them)
[Hint: you’ll need to do long division twice.]
Solution: Since 2 − 3 i is a root of f (x) we know its conjugate 2 + 3i is also a root of f (x) (by the conjugate root theorem).
Similarly we know i and −i are roots. So the relationship between roots and factors give us
Roots Linear Factors
2 − 3 i ←→ x − (2 − 3 i) = x − 2 + 3i 2 + 3i ←→ x − (2 + 3i) = x − 2 − 3 i
i ←→ x − i −i ←→ x + i
Thus we know f (x) = (x−2+3i)(x− 2 − 3 i)(x−i)(x+i)(? ).
Since
(x−2+3i)(x− 2 − 3 i) = x^2 − 2 x− 3 ix− 2 x+4+6i+3ix− 6 i− 9 i^2 = x^2 − 4 x+
and
(x − i)(x + i) = x^2 − ix + ix − i^2 = x^2 + 1
we actually have f (x) = (x^2 − 4 x + 13)(x^2 + 1)(? ). To
find the missing factor (we can either foil (x^2 − 4 x + 13)(x^2 + 1)
and do long division once or) we can do long division twice as
follows:
x^4 − 6 x^3 + 6x^2 + 34x − 195
x^2 + 1
x^6 − 6 x^5 + 7x^4 + 28x^3 − 189 x^2 + 34x − 195 − x^6 − x^4
− 6 x^5 + 6x^4 + 28x^3
6 x^5 + 6x^3
6 x^4 + 34x^3 − 189 x^2
− 6 x^4 − 6 x^2
34 x^3 − 195 x^2 + 34x
− 34 x^3 − 34 x
− 195 x^2 − 195
195 x^2 + 195
0
so f (x) = (x^2 + 1)(x^4 − 6 x^3 + 6x^2 + 34x − 195) and
x^2 − 2 x − 15
x^2 − 4 x + 13
x^4 − 6 x^3 + 6x^2 + 34x − 195 − x^4 + 4x^3 − 13 x^2
− 2 x^3 − 7 x^2 + 34x 2 x^3 − 8 x^2 + 26x
− 15 x^2 + 60x − 195
15 x^2 − 60 x + 195
0
so x^4 − 6 x^3 + 6x^2 + 34x − 195 = (x^2 − 4 x + 13)(x^2 − 2 x − 15)
which implies
f (x) = (x^2 + 1)(x^2 − 4 x + 13)(x^2 − 2 x − 15)
Now x^2 − 2 x − 15 = (x − 5)(x + 3) So we have
f (x) = (x − 2 + 3i)(x − 2 − 3 i)(x − i)(x + i)(x − 5)(x + 3)
