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Law of Cosines Proof Video

Notes

Questions

Solve each triangle

\(\textbf{1)}\) Solve the triangle.

Show Answer

\(\angle A\approx73.9^{\circ} \,\,\, \angle B\approx46.1^{\circ} \,\,\, c=\sqrt{13}\approx3.61\)

Show Work

\(\text{Step 1: Find the missing side, side } c\)

\(\,\,\,\,\,\,c^2=a^2+b^2-2ab\cos{C}\)

\(\,\,\,\,\,\,c^2=4^2+3^2-2(4)(3)\cos{60}\)

\(\,\,\,\,\,\,c^2=4^2+3^2-2(4)(3)\cdot \frac{1}{2}\)

\(\,\,\,\,\,\,c^2=16+9-12\)

\(\,\,\,\,\,\,c^2=13\)

\(\,\,\,\,\,\,c=\sqrt{13}\approx 3.61\)

\(\text{Step 2: Find the first missing angle, } \angle B\)

\(\,\,\,\,\,\,\displaystyle\frac{\sin{C}}{c}=\frac{\sin{B}}{b}\)

\(\,\,\,\,\,\,\displaystyle\frac{\sin{60}}{\sqrt{13}}=\frac{\sin{B}}{3}\)

\(\,\,\,\,\,\,\displaystyle \sqrt{13} \cdot \sin{B}=3 \cdot \sin{60}=\)

\(\,\,\,\,\,\,\displaystyle 3.605551 \cdot \sin{B} =3 \cdot .866025\)

\(\,\,\,\,\,\,\displaystyle 3.605551 \cdot \sin{B} =2.59808\)

\(\,\,\,\,\,\,\displaystyle \sin{B} =.720578\)

\(\,\,\,\,\,\,\displaystyle B=\sin^{-1}(.720578)\)

\(\,\,\,\,\,\,\displaystyle \angle B \approx 46.1^{\circ}\)

\(\text{Step 3: Find the last missing angle, } \angle A\)

\(\,\,\,\,\,\,\angle A+\angle B + \angle C = 180\)

\(\,\,\,\,\,\,\angle A+ 46.1 + 60 = 180\)

\(\,\,\,\,\,\,\angle A+ 106.1 = 180\)

\(\,\,\,\,\,\,\displaystyle \angle A \approx 73.9^{\circ}\)

\(\text{The answer is}\angle A\approx73.9^{\circ} \,\,\, \angle B\approx46.1^{\circ} \,\,\, c=\sqrt{13}\approx3.61\)

\(\textbf{2)}\) Solve the triangle.

Show Answer

\(\angle A\approx106.6^{\circ} \,\,\, \angle B\approx48.2^{\circ} \,\,\, \angle C\approx25.2^{\circ}\)

Show Work

\(\text{Step 1: Find the largest angle, } \angle A\)

\(\,\,\,\,\,\,a^2=b^2+c^2-2bc\cos{A}\)

\(\,\,\,\,\,\,9^2=7^2+4^2-2(7)(4)\cos{A}\)

\(\,\,\,\,\,\,81=49+16-56\cos{A}\)

\(\,\,\,\,\,\,81=65-56\cos{A}\)

\(\,\,\,\,\,\,16=-56\cos{A}\)

\(\,\,\,\,\,\,-.285714=\cos{A}\)

\(\,\,\,\,\,\,\cos^{-1}(-.285714)=A\)

\(\,\,\,\,\,\,\angle A\approx106.6^{\circ}\)

\(\text{Step 2: Find the second largest angle, } \angle B\)

\(\,\,\,\,\,\,\displaystyle\frac{\sin{A}}{a}=\frac{\sin{B}}{b}\)

\(\,\,\,\,\,\,\displaystyle\frac{\sin{106.6}}{9}=\frac{\sin{B}}{7}\)

\(\,\,\,\,\,\,\displaystyle 9\sin{B}=7\sin{106.6}\)

\(\,\,\,\,\,\,\displaystyle 9\sin{B}=7\cdot .95832\)

\(\,\,\,\,\,\,\displaystyle 9\sin{B}=6.7082\)

\(\,\,\,\,\,\,\displaystyle \sin{B}=.745356\)

\(\,\,\,\,\,\,\displaystyle B=\sin^{-1}(.745356)\)

\(\,\,\,\,\,\,\displaystyle \angle B \approx 48.2^{\circ}\)

\(\text{Step 3: Find the last angle, } \angle C\)

\(\,\,\,\,\,\,\angle A+\angle B + \angle C = 180\)

\(\,\,\,\,\,\,106.6+48.2 + \angle C = 180\)

\(\,\,\,\,\,\,154.8 + \angle C = 180\)

\(\,\,\,\,\,\, \angle C = 25.2^{\circ}\)

\(\text{The answer is }\angle A\approx106.6^{\circ} \,\,\, \angle B\approx48.2^{\circ} \,\,\, \angle C\approx25.2^{\circ}\)

\(\textbf{3)}\) Solve the triangle.

Show Answer

\( \angle C \approx 62.2^{\circ} ,\,\angle E \approx 33.5^{\circ} ,\,\angle J \approx 84.3^{\circ} \)

Show Work

\(\text{Step 1: Find the largest angle, } \angle J\)

\(\,\,\,\,\,\,j^2=c^2+e^2-2ce\cos{J}\)

\(\,\,\,\,\,\,9^2=8^2+5^2-2(8)(5)\cos{J}\)

\(\,\,\,\,\,\,81=64+25-80\cos{J}\)

\(\,\,\,\,\,\,81=89-80\cos{J}\)

\(\,\,\,\,\,\,-8=-80\cos{J}\)

\(\,\,\,\,\,\,0.1=\cos{J}\)

\(\,\,\,\,\,\,\cos^{-1}(0.1)=J\)

\(\,\,\,\,\,\,\angle J\approx84.3^{\circ}\)

\(\text{Step 2: Find the second largest angle, } \angle C\)

\(\,\,\,\,\,\,\displaystyle\frac{\sin{J}}{j}=\frac{\sin{C}}{c}\)

\(\,\,\,\,\,\,\displaystyle\frac{\sin{84.3}}{9}=\frac{\sin{C}}{8}\)

\(\,\,\,\,\,\,\displaystyle 9\sin{C}=8\sin{84.3}\)

\(\,\,\,\,\,\,\displaystyle 9\sin{C}=8\cdot 0.995\)

\(\,\,\,\,\,\,\displaystyle 9\sin{C}=7.9604\)

\(\,\,\,\,\,\,\displaystyle \sin{C}=0.88449\)

\(\,\,\,\,\,\,\displaystyle C=\sin^{-1}(0.88449)\)

\(\,\,\,\,\,\,\displaystyle \angle C \approx 62.2^{\circ}\)

\(\text{Step 3: Find the last angle, } \angle E\)

\(\,\,\,\,\,\,\angle J+\angle C + \angle E = 180\)

\(\,\,\,\,\,\,84.3+62.2 + \angle E = 180\)

\(\,\,\,\,\,\,146.5 + \angle E = 180\)

\(\,\,\,\,\,\, \angle E = 33.5^{\circ}\)

\(\textbf{4)}\) Solve the triangle.

Show Answer

\( \angle B \approx 53.1^{\circ} ,\, \angle M \approx 36.9^{\circ} ,\, \angle T \approx 90^{\circ} \)

Show Work

\(\text{Step 1: Find the largest angle, } \angle T\)

\(\,\,\,\,\,\,t^2=m^2+b^2-2mb\cos{T}\)

\(\,\,\,\,\,\,15^2=9^2+12^2-2(9)(12)\cos{T}\)

\(\,\,\,\,\,\,225=81+144-216\cos{T}\)

\(\,\,\,\,\,\,225=225-216\cos{T}\)

\(\,\,\,\,\,\,-0=-216\cos{T}\)

\(\,\,\,\,\,\,0=\cos{T}\)

\(\,\,\,\,\,\,\cos^{-1}(0)=T\)

\(\,\,\,\,\,\,\angle T=90^{\circ}\)

\(\text{Step 2: Find the second largest angle, } \angle B\)

\(\,\,\,\,\,\,\displaystyle\frac{\sin{T}}{t}=\frac{\sin{B}}{b}\)

\(\,\,\,\,\,\,\displaystyle\frac{\sin{90}}{15}=\frac{\sin{B}}{12}\)

\(\,\,\,\,\,\,\displaystyle 15\sin{B}=12\sin{90}\)

\(\,\,\,\,\,\,\displaystyle 15\sin{B}=12\)

\(\,\,\,\,\,\,\displaystyle \sin{B}=0.8\)

\(\,\,\,\,\,\,\displaystyle B=\sin^{-1}(0.8)\)

\(\,\,\,\,\,\,\displaystyle \angle B \approx 53.1^{\circ}\)

\(\text{Step 3: Find the last angle, } \angle M\)

\(\,\,\,\,\,\,\angle T+\angle B + \angle M = 180\)

\(\,\,\,\,\,\,90+53.1 + \angle M = 180\)

\(\,\,\,\,\,\,143.1 + \angle M = 180\)

\(\,\,\,\,\,\, \angle M = 36.9^{\circ}\)

\(\textbf{5)}\) Solve the triangle.

Show Answer

\( \angle D \approx 123^{\circ},\, \angle F \approx 16^{\circ},\, \angle G \approx 41^{\circ} \)

\(\textbf{6)}\) Solve the triangle.

Show Answer

\( \angle A \approx 23^{\circ},\, \angle D \approx 67^{\circ},\, \angle N \approx 90^{\circ} \)

\(\textbf{7)}\) Solve the triangle.

Show Answer

\( a \approx 4.194,\, \angle B \approx 123.01^{\circ}, \, \angle C \approx 33.99^{\circ}\)

\(\textbf{8)}\) Solve the triangle.

Show Answer

\( a \approx 7.522,\, \angle B \approx 87.99^{\circ}, \, \angle C \approx 22.01^{\circ}\)

\(\textbf{9)}\) Solve the triangle.

Show Answer

\( c= \sqrt{7} \approx 2.6458,\, \angle A \approx 19.11^{\circ}, \, \angle B \approx 100.89^{\circ} \)

\(\textbf{10)}\) Solve the triangle.

Show Answer

\( b=7,\, \angle A \approx 21.79^{\circ}, \, \angle C \approx 38.21^{\circ} \)

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\(\bullet\text{ Area of SAS Triangles}\)
\(\,\,\,\,\,\,\,\,\text{Area}=\frac{1}{2}ab \sin{C}\) \(…\)

\(\bullet\text{ Law of Cosines}\)
\(\,\,\,\,\,\,\,\,a^2=b^2+c^2-2bc \cos{A}\) \(…\)

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\(\,\,\,\,\,\,\,\,\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}\) \(…\)

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\(\bullet\text{ Trigonometry-Pythagorean Identities}\)
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\(\bullet\text{ Andymath Homepage}\)

In Summary

The Law of Cosines is a mathematical formula that relates the lengths of the sides of a triangle to the cosine of one of its angles. It is an important tool for solving problems involving triangles, particularly in geometry and trigonometry.

The Law of Cosines states that in any triangle, the square of the length of one side is equal to the sum of the squares of the lengths of the other two sides minus twice the product of those two sides and the cosine of the angle between them. This can be written as:

\(a^2 = b^2 + c^2 – 2bc \cdot \cos{(A)}\)

We learn about the Law of Cosines because it is a useful tool for solving problems involving triangles. It should be used when you want to solve an unknown side of angles in SAS triangles or SSS triangles.

The Law of Cosines is typically taught in a high school geometry class along with Law of Sines. And these lessons occur around the middle to the end of a entire section on trigonometry.

The most common mistake when using the Law of Cosines is to add the \(b^2 + c^2\) and the \(2bc\) together before multiplying the \(2bc\) by the \(\cos{(A)}\). The correct order of operations multiplies the \(2bc \cdot \cos{(A)}\) and then adding that to \(b^2 + c^2\).

It should be noted that the pythagorean theorem is a specific case of the law of cosines. If the angle between the two sides is 90 degrees, then the \(2bc \cdot \cos{(A)}\) is equal to zero and we end up with simply \(a^2+b^2=c^2\).

Real world examples of Law of Cosines

Finding the distance between two points on a map: The Law of Cosines can be used to find the distance between two points on a map. If you can treat the points as the vertices of an SAS triangle, the distance between the two points can be calculated using the Law of Cosines.

Calculating the height of a building: The Law of Cosines can be used to find the height of a building if enough info is know to create an SAS triangle with the third side being the building.

Measuring the height of a mountain: Same idea as the height of a building, as long as an SAS triangle can be formed with the height of the mountain as the third unknown side, we can use law of cosines to find the height of the building.

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