The Law of Cosines

We will discuss here about the law of cosines or the cosine rule which is required for solving the problems on triangle. 

In any triangle ABC, Prove that,

(i) b\(^{2}\) = c\(^{2}\) + a\(^{2}\) - 2ca. cos B or, cos B =  \(\frac{c^{2} + a^{2} - b^{2}}{2ca}\)

(ii) a\(^{2}\) = b\(^{2}\) + c\(^{2}\) - 2ab. cos A or, cos A = \(\frac{b^{2} + c^{2} - a^{2}}{2bc}\)

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

 

Proof of the law of cosines:

Let ABC is a triangle. Then the following three cases arise:

Case I: When the triangle ABC is acute-angled:

Now form the triangle ABD, we have,

cos B = BD/BC

⇒ cos B = BD/c

⇒ BD = c cos B ……………………………………. (1)

Again from the triangle ACD, we have

cos C = CD/CA

⇒ cos C = CD/b

⇒ CD = b cos C

By using the Pythagoras theorem on the triangle ACD, we get

AC\(^{2}\) = AD\(^{2}\) + CD\(^{2}\)

⇒ AC\(^{2}\) = AD\(^{2}\) + (BC - BD)\(^{2}\)

⇒ AC\(^{2}\) = AD\(^{2}\) + BC\(^{2}\) + BD\(^{2}\) - 2 BC ∙ BD

⇒ AC\(^{2}\) = BC\(^{2}\) + (AD\(^{2}\) + BD\(^{2}\)) - 2 BC ∙ BD

⇒ AC\(^{2}\) = BC\(^{2}\) + AB\(^{2}\) - 2 BC ∙ BD, [Since From triangle, we get, AD\(^{2}\) + BD\(^{2}\) = AB\(^{2}\)]

⇒ b\(^{2}\) = a\(^{2}\) + c\(^{2}\) - 2a ∙ c cos B, [From (1)]

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


Case II: When the triangle ABC is obtuse-angled:

The triangle ABC is obtuse angled.

Now, draw AD from A which is perpendicular to produced BC. Clearly, D lies on produced BC.

Now from the triangle ABD, we have,

cos (180° - B) = BD/AB

⇒- cos B = BD/AB, [Since, cos (180° - B) = - cos B]

⇒ BD = -AB cos B

⇒ BD = -c cos B ……………………………………. (2)

By using the Pythagoras theorem on the triangle ACD, we get

AC\(^{2}\) = AD\(^{2}\) + CD\(^{2}\)

⇒ AC\(^{2}\) = AD\(^{2}\) + (BC + BD)\(^{2}\)

⇒ AC\(^{2}\) = AD\(^{2}\) + BC\(^{2}\) + BD\(^{2}\) + 2 BC ∙ BD

⇒ AC\(^{2}\)= BC\(^{2}\)+ (AD^2 + BD^2) + 2 BC ∙ BD

⇒ AC\(^{2}\) = BC\(^{2}\) + AB\(^{2}\) + 2 BC ∙ BD, [Since From triangle, we get, AD\(^{2}\) + BD\(^{2}\) = AB\(^{2}\)]

⇒ b\(^{2}\) = a\(^{2}\) + c\(^{2}\) + 2a ∙ (-c - cos B), [From (2)]

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


Case III: Right angled triangle (one angle is right angle):  The triangle ABC is right angled. The angle B is a right angle.

Now by using the Pythagoras theorem we get,

b\(^{2}\) = AC\(^{2}\) = BC\(^{2}\) + BA\(^{2}\) = a\(^{2}\) + c\(^{2}\)

⇒ b\(^{2}\) = a\(^{2}\) + c\(^{2}\)

⇒ b\(^{2}\) = a\(^{2}\) + c\(^{2}\) - 2ac cos B, [We know that cos 90° = 0 and B = 90°. Therefore, cos B = 0] or, cos B = \(\frac{c^{2} + a^{2} - b^{2}}{2ca}\)

Therefore, in all three cases, we get,

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

Similarly, we can prove that the formulae (ii) a\(^{2}\) = b\(^{2}\) + c\(^{2}\) - 2ab. cos A or, cos A = \(\frac{b^{2} + c^{2} - a^{2}}{2bc}\) and (iii) c\(^{2}\) = a\(^{2}\) + b\(^{2}\) - 2ab. cos C or, cos C = \(\frac{a^{2} + b^{2} - c^{2}}{2ab}\).


Solved problem using the law of Cosines:

In the triangle ABC, if a = 5, b = 7 and c = 3; find the angle B and the circum-radius R.

Solution:

Using the formula, cos B = \(\frac{c^{2} + a^{2} - b^{2}}{2ca}\) we get,

cos B = \(\frac{3^{2} + 5^{2} - 7^{2}}{2 ∙ 3 ∙ 5}\)

cos B = \(\frac{9 + 25 - 49}{30}\)

cos B = - 1/2

cos B = cos 120°

Therefore, B = 120°

Again, if R be the required circum-radius then,

b/sin B = 2R

⇒ 2R = 7/sin 120°

⇒ 2R = 7 ∙ 2/√3                      

Therefore, R = 7/√3 = (7√3)/3 units.

 Properties of Triangles






11 and 12 Grade Math 

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