Describe how the Law of Cosines can be used to solve the ambiguous case of the oblique triangle ABC, where a = 12 feet, b = 30 feet, and A = 20°. Is the result the same as when the Law of Sines is used to solve the triangle? Describe the advantages and the disadvantages of each method.

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01-10-2019
Mathematics

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Describe how the Law of Cosines can be used to solve the ambiguous case of the oblique triangle ABC, where a = 12 feet, b = 30 feet, and A = 20°. Is the result the same as when the Law of Sines is used to solve the triangle? Describe the advantages and the disadvantages of each method.

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jonathanjmc30 jonathanjmc30 · Answer 1 · 2019-10-06T07:05:56+02:00

Answer:

First question: Sure, we can use the law of cosines to solve the oblique triangle solving a quadratic equation.

Second question: Yes, it is, but is easier to use the law of sines than cosine law due the simplicity of the process with sines law.

Step-by-step explanation:

First of all we have to consider that in our triangle we don't have the side c and angles B and C, (remember that the notation used in trigonometry for triangles is in such a way that the angles and opposites sides have the same letters but the first one in capital letter and the later in small letters), so we have, using the law of cosines:

[tex]12^{2}=c^{2} +30^{2}-2(30)c cos(20)[/tex]

then, simplifying and leaving in the quadratic general form of [tex]ax^{2} +bx+c=0[/tex] we get in the variable c instead x:

[tex]c^{2}-56.381c+756=0

Then, solving this equatric equation by the quadratic formula [tex]{c= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}[/tex] with a=1, b=-56.381 and c=756, we get:

c=21.9692 and c=34.41.

Using the law of sine:

We have the following relation:

[tex]\frac{\sin(20)}{12}=\frac{\sin(B)}{30}[/tex]

Solving for [tex]\sin(B)[\tex] we get [tex]\sin(B)=0.85[\tex], and using the inverse function for sine we have that [tex]B=\sin^{-1}(0.85)=58.765[/tex]

Then, by the sum of the angles inside of a triangle we have that C=101.234°.

Finally, using again the law of sines we have:

[tex]\frac{\sin(101.234)}{c}=\frac{\sin(20)}{12}[/tex]

and solving for c we have that

[tex]c=34.41[/tex]

In conclusion: we got the same result using sine or cosine law, but clearly is easier to use the law of sine because using law of cosine we have to solve a quadratic equation in the first step to solve the triangle via cosine law.