Download Name: Law of Sines and Cosines Directions: For the following

Name:
Law of Sines and Cosines
Directions: For the following worksheet, please put only your answers in the spaces shown below. Please remember to
give exact answers, not decimal approximations. Attach scratch paper to show your work.
Remember that SOHCAHTOA only works when you have a right triangle. Today, we’ll consider how to compute side
lengths of triangles that are not right triangles. We start by recapping some properties of triangles:
1.
2.
3.
4.
The interior angles of all triangles add up to 180◦ .
The smallest angle will be opposite the smallest side.
The largest angle will be opposite the largest side.
If two angles are the same in magnitude, then their opposite sides will be the same lengths.
Law of Sines: The Law of Sines is used when you are given an angle, it’s opposite side,
and something else. The Law of Sines equation is:
sin(α)
sin(β)
sin(θ)
=
=
a
b
c
NOTE: In this case c does not necessarily have to be the longest side.
Sample Problem: Determine the length of QR for the triangle shown to the left.
Using the 180◦ property, we know that the remaining side is 65◦ . Notice that we have an
angle (65◦ ), its opposite side (8), and something else, so we can use the Law of Sines:
sin(65◦ )
8
=
sin(68◦ )
PQ
sin(47◦ )
QR
=
When working with the Law of Sines, you only do two equations at a time, so ignore the
middle part, then cross multiply.
sin(65◦ )
sin(47◦ )
=
8
QR
−→
QR · sin(65◦ ) = 8 sin(47◦ ).
Thus, QR =
8 sin(47◦ )
≈ 6.4557.
sin(65◦ )
Notice that QR is opposite the smallest angle, so it makes sense that this side is smaller than 8.
Examples: Use the Law of Sines to compute the following:
1. Using the same triangle as above, compute P Q.
sin(65◦ )
sin(68◦ )
=
8
PQ
−→
P Q sin(65◦ ) = 8 sin(68◦ )
−→
PQ =
8 sin(68◦ )
sin(65◦ )
This approximates to 8.18, which makes sense since this side is opposite the largest angle.
2. A triangle has base angles 50◦ and 60◦ and a base of length 11. Determine the length of the shortest side of the
triangle.
Using the 180 property, the missing angle is 70◦ . And using Law of Sines, we get:
sin(70◦ )
sin(50◦ )
=
11
x
−→
P Q sin(70◦ ) = 11 sin(50◦ )
−→
PQ =
11 sin(50◦ )
sin(70◦ )
This approximates to 8.97, which makes sense since this side is opposite the smallest angle.
3. A triangle has interior angles 43◦ , 66◦ , and 71◦ . Its longest side has length 23. Answer the following:
(a) Determine the perimeter of the triangle.
To find the perimeter, you need to compute the other sides. Use Law of Sines to get:
a=
23 sin(66◦ )
sin(71◦ )
and
c=
23 sin(43◦ )
sin(71◦ )
23 sin(66◦ )
23 sin(43◦ )
+
sin(71◦ )
sin(71◦ )
(b) Determine the area of the triangle. (Use the formula from Section 5.7.)
Thus, the perimeter is P = 23 +
Using b = 23 and θ = 43◦ , we have: A =
1
1 23 sin(66◦ )
· a · b · sin(θ) = ·
· 23 · sin(43◦ )
2
2
sin(71◦ )
Law of Cosines: The Law of Cosines is used when the Law of Sines does not apply. The Law of Cosines equation is:
c2
=
a2 + b2 − 2 · a · b cos(θ)
Note: You may have to label your triangle carefully to use this formula. For this equation, c is always opposite of θ, and
a and b are the side lengths touching θ.
Sample Problem: Determine the length of QR for the triangle shown to the left.
Notice that we are given one angle (47◦ ), but we do not know its opposite side. Thus, we
cannot use Law of Sines. Using the Law of Cosines, if θ = 47◦ , then c = QR, and:
c2 = 82 + 132 − 2(8)(13) cos(47◦ )
p
And if we take a square root, we have c = 82 + 132 − 2(8)(13) cos(47◦ ).
Examples: Use the Law of Cosines to compute the following.
4. An obtuse triangle has an interior angle of 114◦ , and its smallest two sides are of length 13 and 15. Determine the
perimeter of the triangle.
You’ll need to be able to come up with this diagram. Using Law of
Cosines, you’ll get:
p
c = 132 + 152 − 2(13)(15) cos(114◦ )
5. The area of the triangle shown below is 100. Answer the following:
(a) Solve for a.
100 =
1
· a · 20 · sin(55◦ )
2
−→
10
=a
sin(55◦ )
(b) Determine the base of the triangle.
Using Law of Cosines, we first note that θ = 55◦ and its opposite side (the base) is c.
s
2
10
10
2−2
c=
+
20
(20) cos(55◦ )
sin(55◦ )
sin(55◦ )