Ever faced a triangle that isn't a neat right triangle and wondered how to crack it? Enter the Cosine Rule—also known as the Law of Cosines—your go-to formula for those trickier triangles! This rule links the lengths of a triangle’s sides to the cosine of one of its angles. It's like a leveled-up Pythagorean theorem that works even when there’s no 90° angle in sight.

In simple terms, the Cosine Rule helps you:

• Calculate the length of any side if you know two sides and the angle between them.
• Find an angle if you know all three sides.

Whether you're prepping for the SAT, ACT, PSAT/NMSQT, GED, GRE, GMAT, AP Exams, PERT, Accuplacer, or even the MCAT, knowing how to use the Cosine Rule gives you a real edge!

Cosine Rule

Cosine Rule is a rule that relates two sides of a triangle and the angle between them. This is used to find either any unknown angle or any unknown side.

The Cosine Rule states that “the square of the length of any side of a given triangle is equal to the sum of the squares of the length of the other sides minus twice the product of the other two sides multiplied by the cosine of angle included between them”. This rule is also called the Law of Cosine.

If we have \( \triangle ABC \) as shown above, then according to the Cosine Rule we can write the formula as given below.

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

This formula can also be written in other ways depending upon the sides and angles given and when we want to find an unknown side

\( b^2=a^2+c^2-2ac\cos{B} \), or

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

When we have been given the sides and we want to find the angle then we rearrange the formula as shown below.

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

\( \cos B=\frac{a^2+c^2-b^2}{2ac} \), and

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

Proof of Cosine Rule

The proof of the Cosine Rule is provided below.

Let ABC be a triangle given below.

In the right angled triangle BCD, from the definition of cosine we get

\( \cos C=\frac{CD}{a}\Rightarrow CD=a\cos C \)

We get, \( DA=b-\cos C \) … (i)

Also from the \( \triangle BCD \) from the definition of sine we get

\( \sin C=\frac{BD}{a}\Rightarrow BD=a\sin C \) … (ii)

In \( \triangle ABD \) applying the Pythagorean Theorem we get,

\( c^2=BD^2+DA^2 \)

Substituting for BD and DA from (2) and (3) we get,

\( c^2=(a\sin C)^2+(b−a\cos C)^2\Rightarrow c^2=a^2\sin^2C+b^2−2ab\cos C+a^2\cos^2C\Rightarrow c^2=a^2\sin^2C+a^2\cos^2C+b^2−2ab\cos C \)

Factoring out \( a^2 \) we get

\( c^2=a^2(\sin^2C+\cos^2C)+b^2−2ab\cos C \)

We also known that \( \sin^2\theta+\cos^2\theta=1 \)

Therefore we get,

\( c^2=a^2+b^2−2ab\cos C \)

Hence proved.

Derivation of Cosine Formula from Law of Sines

Let us consider a triangle with sides as a, b, and c, and their respective angles as x, y and z, as shown above.

From the Law of Sines we can write the following.

\( \frac{c}{\sin z}=\frac{b}{\sin y}=\frac{a}{\sin x} \) …(i)

We also know that the sum of the interior angles of a triangle is \( 180\degree \).

\( \therefore x+y+z=180\degree \) …(ii)

Now using the third equation system, we get

\( \frac{c}{\sin z}=\frac{b}{\sin(x+z)} \) …(iii)

\( \Rightarrow\frac{c}{\sin z}=\frac{a}{\sin x} \)

Next dividing the whole equation by \( \cos z \) we get,

\( c(\sin x+\tan z\cos x)=b\tan z \)

\( \Rightarrow\frac{c\sin x}{\cos z}=a\tan z \)

\( \Rightarrow\frac{c^2\sin^2x}{\cos^2z}=\tan z \)

From equation (iii we get,

\( \frac{c\sin x}{b}-c\cos x=\tan z \)

\( \Rightarrow1+\tan^2z=\frac{1}{\cos^2z} \)

\( \Rightarrow c^2\si)^2x\left(1+\frac{c^2\sin^2x}{(b-c\cos x)^2}\right)=a^2\left(\frac{c^2\sin^2x}{(b-c\cos x)^2}\right) \)

Finally multiplying the equation by \( (b-c\cos x)^2 \) and rearranging it we get the required Cosine Rule.

\( a^2=b^2+c^2-2bc\cos x \)

Hence Proved.

Cosine Rule Solved Example

Problem 1:

In \( \triangle ABC \), AB = 42cm, BC = 37cm and AC = 26cm. Calculate the value of all the three interior angles.

Solution:

In the above image we have labeled the triangle with the given lengths.

We will find the interior angles using the rearranged form of the Cosine Rule.

\( cos\left(A\right)=\frac{b^2+c^2-a^2}{2bc}=\frac{\left(26\right)^2+\left(42\right)^2-\left(37\right)^2}{2\times26\times42}=\frac{1071}{2184}=0.4904 \)

\( \Rightarrow A=\cos^{-1}\left(0.4904\right)=60.63\degree \)

Similarly we get ,

\(

\cos\left(B\right)=\frac{a^2+c^2-b^2}{2ac}=\frac{\left(37\right)^2+\left(42\right)^2-\left(26\right)^2}{2\times37\times42}=\frac{2457}{3108}=0.7905 \)

\( \Rightarrow B=\cos^{-1}\left(0.7905\right)=37.76\degree \)

And also,

\( \cos\left(C\right)=\frac{a^2+b^2-c^2}{2ab}=\frac{\left(37\right)^2+\left(26\right)^2-\left(42\right)^2}{2\times37\times26}=\frac{281}{1924}=0.1460 \)

\( \Rightarrow C=\cos^{-1}\left(0.1460\right)=81.61\degree \)

Thus we get \( \angle A=60.63\degree,\ \angle B=37.76\degree,\ \angle C=81.61\degree \) as the three interior angles.

Problem 2:

Calculate the third unknown side of \( \angle ABC \). Given \( \angle A = 29\degree \), AB = 76cm and AC = 105cm.

Solution:

In the above image we have labeled the triangle with the given parameters, and we need to find the lengths of BC.

As we have the lengths of two sides and the angle between them, so we can use the Cosine rule formula to get the unknown value.

Therefore,

\( a^2=b^2+c^2-2bc\cos A=\left(105\right)^2+\left(76\right)^2-\left(2\times105\times76\times\cos29\degree\right)=16801-13,958.9305=2,842.0695 \)

\( a=\sqrt{2,842.0695}=53.311 \)

Thus the value of the third side of the triangle is 53.311cm.

Conclusion

The Cosine Rule is a powerful tool for handling non-right triangles, making it essential for exam success. Whether you need to find missing sides or angles, this formula simplifies the process and boosts your problem-solving skills. Perfect for tests like the SAT, ACT, GRE, GMAT, AP Exams, and MCAT, mastering the Cosine Rule can give you the confidence to tackle even the trickiest trigonometry problems. Keep practicing and sharpen your skills!