Download The Laws of SINES

The angle of elevation of the
top of a building measured
from point A is 25o. At point
D which is 15m closer to the
building, the angle of
elevation is 35o Calculate the
height of the building.
T
10o
36.5
35o
B
Angle TDA = 180 – 35 = 145o
Angle DTA = 180 – 170 = 10o
TD
15

Sin 25o Sin10o
15Sin 25o
TD 
 36.5 m
Sin10
145o
25o
D
15 m
Sin 35o 
A
TB
36.5
 TB  36.5Sin 35o  20.9 m
The angle of elevation of the top of a column measured from point A, is 20o.
The angle of elevation of the top of the statue is 25o. Find the height of the
statue when the measurements are taken 50 m from its base
Angle BCA = 180 – 110 = 70o
Cos 20o 
50
Angle ACT = 180 – 70 = 110o
TC
53.21

Sin 5o Sin 65o
AC
50
53.21 Sin 5
 5.1 m (1dp )
Cos 20o  TC 
Sin 65o
 53.21 m (2dp )
 AC 
Angle ATC =
T
180 – 115 = 65o
65o
110o
C
70o
5o
A
20o
25o
50 m
B
A fishing boat leaves a harbour (H) and travels due East for 40 miles to a
marker buoy (B). At B the boat turns left and sails for 24 miles to a
lighthouse (L). It then returns to harbour, a distance of 57 miles.
(a) Make a sketch of the journey.
(b) Find the bearing of the lighthouse from the harbour. (nearest degree)
572  402  24 2
CosA 
2x 57x 40
A  20.4o
L
 Bearing  90  20.4  070o
57 miles
H
24 miles
A
40 miles
B
An AWACS aircraft takes off from RAF
Waddington (W) on a navigation
exercise. It flies 530 miles North to
a point (P) as shown, It then turns
left and flies to a point (Q), 670
miles away. Finally it flies back to
base, a distance of 520 miles.
Find the bearing of Q from point P.
b2 c 2 a2
CosA 
2bc
5302  6702  5202
CosP 
2x 530x 670
P  48.7o
 Bearing  180  48.7  229o
Not to Scale
P
670 miles
530 miles
Q
520 miles
W
Complete Table
Known
AAA
AAS
ASA
SSA
SAS
SSS
Diagram
Notes
The Law
of
COSINES
The Law of Cosines
Use to find SIDES
Use to find ANGLES
𝒂𝟐 = 𝒃𝟐 + 𝒄𝟐 − 𝟐𝒃𝒄𝑪𝒐𝒔𝑨
𝒃𝟐 + 𝒄𝟐 − 𝒂𝟐
𝒄𝒐𝒔𝑨 =
𝟐𝒃𝒄
The Law of Cosines
Use to find SIDES
𝒂𝟐 = 𝒃𝟐 + 𝒄𝟐 − 𝟐𝒃𝒄𝑪𝒐𝒔𝑨
𝒂=
𝒃𝟐 + 𝒄𝟐 − 𝟐𝐛𝐜𝐂𝐨𝐬𝐀
The Law of Cosines
Use to find ANGLES
𝒃𝟐 + 𝒄𝟐 − 𝒂𝟐
𝒄𝒐𝒔𝑨 =
𝟐𝒃𝒄
𝐴=
𝟐 +𝒄𝟐 −𝒂𝟐
𝒃
𝐶𝑜𝑠 −1 (
)
𝟐𝒃𝒄
Rewrite with different labels:
The Law of COSINES
For any triangle (right, acute or obtuse), you may
use the following formula to solve for missing
sides or angles:
2
2
2
2
2
2
a  b  c  2bc cos A
b  a  c  2accos B
c  a  b  2abcos C
2
2
2
Use Law of COSINES when ...
you have 3 dimensions of a triangle and you need to find
the other 3 dimensions . They cannot be just ANY 3
dimensions though, or you won’t have enough
information to solve the Law of Cosines equations. Use
the Law of Cosines if you are given:


SAS - 2 sides and their included angle
SSS
While you wait:

Solve the following triangles:
Law of Cosines Day 2
Example 1: Given SAS
Solve triangle ABC, given that angle B = 98°, side a = 13
and side c = 20.
First draw a diagram.
b 2  a 2  c 2  2accos B
b 2  132  20 2  2 13 20cos 98
B
98°
C
c = 20
b
A
b 2  641.37
b  25.3
Example 1: Given SAS
Now we have to find angles A and C.
Let’s take on angle A first.
B
a = 13
C
98°
c = 20
b = 25.3
In order to find angle A
should we use?
a) Law of Sines
b) Law of Cosines
A
c) Either law will work
d) Neither will work
LOS vs LOC
SSA… might result in two possible
solutions.
 But not in this case, since there is
already an obtuse angle.
 If an angle might be obtuse, never use the
Law of Sine equation to find it.
 Law of cosines is a better option.

Example 1: Given SAS
Now that we know B and b, we can use the Law of
Sines to find one of the missing angles:
B
a = 13
98°
C
c = 20
b = 25.3
Solution:
b = 25.3, C = 51.5°, A = 30.6°
𝟐 + 𝟐𝟎𝟐 − 𝟏𝟑𝟐
𝟐𝟓.
𝟑
𝑨 = 𝒄𝒐𝒔−𝟏 (
)
𝟐 ∙ 𝟐𝟓. 𝟑 ∙ 𝟐𝟎
A
𝑨~𝟑𝟎. 𝟔°
Example 1: Given SAS
Now that we know B and b, we can use the Law of
Sines to find one of the missing angles:
B
a = 13
98°
C
c = 20
b = 25.3
Solution:
b = 25.3, C = 51.5°, A = 30.5°
25.3
20

sin 98 sin C
1 20sin 98
C  sin 
25.3 
A
C  51.5
A  180  98  51.5  30.5
Example 2: Given SAS
Solve triangle, ABC, given that angle A = 39°, side b =
20 and side c = 15.
Use the Law of Cosines equation that uses b, c and A to find
side a:
a 2  b 2  c 2  2bc cos A
B
a
c = 15
A
39°
a 2  20 2  152  2  20 15cos 39
b = 20
C
a 2  158.71
a  12.6
Example 2: Given SAS
Use the Law of Sines to find one of the missing angles:
B
a = 12.6
c = 15
A
39°
b = 20
C
12.6
15

sin 39 sin C
1 15sin 39 
C  sin 
12.6 
C  48.5
B  180  39  48.5  92.5
Important: Notice that we used the Law of Sine equation to find angle C
rather than angle B. The Law of Sine equation will never produce an obtuse
angle. If we had used the Law of Sine equation to find angle B we would
have gotten 87.5°, which is not correct, it is the reference angle for the
correct answer, 92.5°. If an angle might be obtuse, never use the Law of
Sine equation to find it.
Example 3: Given SSS
Solve triangle, ABC, given that side a = 30, side b = 20 and
side c = 15.
We can use any of the Law of
Cosine equations, filling in a, b
& c and solving for one angle.
A
c = 15
B
b = 20
a = 30
C
Once we have an angle, we
can either use another Law of
Cosine equation to find
another angle, or use the Law
of Sines to find another angle.
Example 3: Given SSS
Important: The Law of Sines will never produce an obtuse angle. If
an angle might be obtuse, never use the Law of Sines to find it. For
this reason, we will use the Law of Cosines to find the largest angle
first (in case it happens to be obtuse).
Angle A is largest because side a is largest:
30 2  202  152  2 20 15cos A
A
c = 15
B
a 2  b 2  c 2  2bc cos A
900  400  225  600cos A
275  600 cos A
b = 20
a = 30
C
275
 cos A
600
1  275 
A  cos
 117.3
600 
Example 3: Given SSS
A
c = 15
B
117.3°
a = 30
Solution:
A = 117.3°
B = 36.3°
C = 26.4°
Use Law of Sines to find angle B or C
(its safe because they cannot be obtuse):
b = 20
C
30
20

sin117.3 sin B
1 20sin 117.3
B  sin 

30
B  36.3
C  180  117.3  36.3  26.4
The Law of Cosines
2
2
2
2
2
2
a  b  c  2bc cos A
b  a  c  2accos B
c 2  a 2  b 2  2abcos C
When given one of these dimension
combinations, use the Law of Cosines
to find one missing dimension and
then use Law of Sines to find the rest.

SAS
 SSS
Important: The Law of Sines will never produce an obtuse angle. If
an angle might be obtuse, never use the Law of Sines to find it.
 Find all the angles created between each pair of cities.
 If the average speed is 49.8 mph, how long will the total trip
take.
Repeat the process with 3 cities
of your choice.
1)
2)
3)
Choose 3 cities or locations.
Sketch a careful map of the three
locations. Find the distance
between each pair of cities…
include these values on your
sketch.
Then find all the angles.
Homework
1)
2)
3)
Complete the triangle table;
Three Cities!
Sec 9.4 page 352 #11, 13, 15-18 all
The above problems are suggested, do
more if you need more practice.
Additional Resources
 Web

Links:
http://oakroadsystems.com/twt/solving.htm#SineLaw
 http://oakroadsystems.com/twt/solving.htm#Detective