Assignment Solution: Laws of Sines and Cosines - Example Problems

This document provides a detailed solution to an assignment focusing on the Law of Sines and the Law of Cosines. It begins by outlining the sine rule and the cosine rule, including when each can be applied, with example problems demonstrating their application. The solution covers calculations for angles and sides of triangles, illustrating the use of these rules in various scenarios. The assignment also addresses the ambiguous case of the Law of Sines, where two possible solutions can exist, providing clear explanations and worked examples. The solution includes references to relevant mathematical resources, ensuring a comprehensive understanding of the concepts. This assignment is designed to help students understand and apply these fundamental trigonometric principles.

Assessment 6.02b: Applying the Laws of Sines and Cosines.
a) The law of sines
The sine rule is given in equation form as follows:
a
sinA = b
sinB= c
sinC
The sine rule can only be used when:
i) At least two angles and one side are provided.
ii) Two sides and any angle are provided
Example
Given that a = 15 cm, angle ABC =50◦ and angle BAC = 70◦
Solution
Using sine rule: 15
sin(70)= b
sin (50)= c
sinC
b = 15 sin (50)
sin ( 70 ) = 12.22 cm
c = 15 sin (60)
sin ( 70 ) =13.82 cm
b) The law of cosines
In equation form, the law is stated as follows

a2 = b2 + c2 − 2bc cosA or,
b2 = a2 + c2 − 2ac cosB or,
c2 = a2 + b2 − 2ab cosC
The cosine rule can only be used when:
i) The three triangle sides are provided.
ii) The included angle and at least two triangle sides are provided.
Example
Given that angle ABC = 50◦, a = 10 cm, c = 7 cm, solve the triangle for the missing angles
and side.
Solution
Using the cosine rule: b2 = a2 + c 2 – 2ac cos B
b2 = 102 + 72 – 2×10×7cos 50
Therefore b2 = 59 cm, = 7.68 cm
Applying sine rule: 7.68
sin(50) = 10
sinA sin A = 0.9974 and thus angle BAC = 85.91◦ and angle
BCA = 180- (85.91 + 50) = 44.09◦
c) The law of sines with the ambiguous case.
The ambiguous case arises when it is possible to have two different angles from the given
data (2 sides and 1 angle). In this case, there are three possible results:
i) One triangle (if one of the given side equals the triangle’s height).
ii) Two different triangles (if one of the sides is greater than the height of the triangle but
less than the other side)
iii) No triangle fits the given information (if the given side is less than the triangle’s
height).
Example
If a = 36 cm, b = 30 cm and angle ABC = 36 degrees:
From the sine rule, 30
sin(50) = 36
sinA therefore, sin A = 0.9192 and thus A = 66.82◦
If A = 66.82◦ and B = 36◦
Then C = 180◦ − 66.82◦ − 36◦ = 77.18◦
If A = 113.18◦ and B = 36◦

Then C = 180◦ − 113.18◦ − 36◦ = 30.82◦
References
[1] Bird, J. O. (2007). Engineering Mathematics (5th ed.). Amsterdam, Netherlands: Elsevier.
[2] Retrieved from http://www.mathguide.com/lessons/LawSines.html